= Quadratic Equation Calculator

Quadratic Equation Calculator

Solve linear, quadratic, cubic, and polynomial equations instantly. Get exact roots, discriminant analysis, quadratic formula steps, factoring, and verification.

Step-by-step Exact roots Discriminant Verification
=
Enter equation
Auto-solving enabled
Math keypad
Use ^ for powers. Enter as equation (with =) or expression (= 0 assumed).
=
Solution
Roots
Answer
Enter an equation above to solve.
Waiting for input.
Type
-
Degree
-
Roots
-

Understanding Quadratic Equations

The word quadratic comes from the Latin quadratus, meaning squared. A quadratic equation is any polynomial equation where the highest power of the variable is exactly 2, written in standard form as ax² + bx + c = 0, where a, b, and c are real constants and a ≠ 0. The presence of the squared term is what gives a quadratic its defining properties: up to two real roots, a symmetric curved graph, and a rich connection between the coefficients and the roots.

Why degree 2 matters

The degree determines the maximum number of roots and the shape of the associated graph. A degree-1 (linear) equation has exactly one root and a straight-line graph. Moving to degree 2 introduces curvature, up to two solutions, and a relationship between the equation's coefficients and its roots captured by Vieta's formulas. The quadratic is the simplest nonlinear polynomial, which is why it appears across physics, engineering, economics, and computer science as the natural step beyond linearity.

Why the graph is a parabola

When you plot y = ax² + bx + c, the squared term creates a symmetric relationship: for any horizontal deviation r from the axis of symmetry, the points x = h + r and x = h − r produce exactly the same y-value. This bilateral symmetry defines a parabola. The parabola opens upward when a > 0 and downward when a < 0. Larger values of |a| produce a narrower, steeper curve; smaller values produce a wider, more gradual curve. The roots of the equation are the x-coordinates where the parabola crosses the x-axis, directly linking the algebraic and geometric perspectives.

What a, b, and c mean

a (leading coefficient): Controls the direction and width of the parabola. Positive a means the parabola opens upward with a minimum vertex; negative a means it opens downward with a maximum vertex. If a = 0, the equation degenerates into a linear equation and is no longer quadratic.

b (linear coefficient): Influences the horizontal position of the axis of symmetry, located at x = −b / (2a). Changing b shifts the vertex left or right while the shape determined by a stays constant.

c (constant term): The y-intercept. When x = 0, the equation gives y = c. The parabola always passes through the point (0, c) regardless of the values of a and b.

Why a cannot be zero

If a = 0, the x² term disappears and the equation becomes bx + c = 0, which is linear and has at most one solution. The quadratic formula, the discriminant, and the parabolic graph all become undefined or degenerate when a = 0. The requirement a ≠ 0 is fundamental, not a minor technicality.

Standard form, factored form, and vertex form

Standard form: ax² + bx + c = 0. Best for applying the quadratic formula and computing the discriminant. This is the starting point for most analytical work.

Factored form: a(x − r1)(x − r2) = 0. Best when the roots are known or easy to identify. The roots r1 and r2 are visible at a glance, and expanding the brackets verifies the coefficients.

Vertex form: a(x − h)² + k = 0. Best for reading off the vertex (h, k) and the axis of symmetry x = h. Useful in graphing, optimization problems, and completing-the-square derivations. Each form reveals different information, and fluency in converting between them is a mark of genuine algebraic understanding.

The Discriminant Explained

The discriminant is defined as Δ = b² − 4ac. It is the expression under the square root in the quadratic formula x = (−b ± √Δ) / 2a. Its sign alone determines the nature and count of the roots before any roots are computed, making it the fastest possible diagnostic for a quadratic equation.

Why Δ > 0 gives two distinct real roots

When the discriminant is positive, √Δ is a real, positive number. The quadratic formula produces two different numerators, (−b + √Δ) and (−b − √Δ), giving two distinct real roots x1 = (−b + √Δ) / 2a and x2 = (−b − √Δ) / 2a. Geometrically, the parabola crosses the x-axis at two separate points. These two points are equidistant from the axis of symmetry x = −b / 2a, which is why the axis lies exactly midway between the roots.

Why Δ = 0 gives one repeated root

When the discriminant is exactly zero, √Δ = 0 and both cases of the ± produce the same value: x = −b / 2a. This is called a repeated root or double root, counted with multiplicity 2 by the Fundamental Theorem of Algebra. Geometrically, the vertex of the parabola sits precisely on the x-axis. The curve touches the axis at that single point without crossing through it. The quadratic factors cleanly as a(x − r)² = 0 where r = −b / 2a.

Why Δ < 0 gives complex roots

When the discriminant is negative, no real number squared gives a negative result. The square root in the quadratic formula becomes imaginary, and the two roots take the form x = (−b ± i√|Δ|) / 2a, where i = √(−1). These are complex conjugate pairs: if one root is p + qi, the other is p − qi. They are mathematically valid solutions but lie outside the real number line. The parabola does not intersect the x-axis at any real-valued point. Saying "no solution" is incorrect; the precise statement is "no real solution."

How the discriminant connects algebra to the graph

The vertex of the parabola sits at height k = c − b² / (4a) above the x-axis. The discriminant satisfies Δ = −4ak. When a > 0 (opens upward) and the vertex is below the axis (k < 0), we get Δ = −4a × k > 0, predicting two crossings. When k = 0, the vertex is on the axis and Δ = 0. When k > 0 with a > 0, the parabola floats entirely above the axis and Δ < 0. The discriminant is, in this sense, a signed measure of the geometric gap between the parabola's turning point and the x-axis.

Predicting x-axis intersections before solving

Compute b² − 4ac using only arithmetic. A positive result guarantees two real x-intercepts. Zero guarantees exactly one (the vertex touching the axis). A negative result guarantees the parabola does not touch the axis. This prediction requires no square roots and is typically far faster than completing a full solution, making the discriminant the ideal first check before investing effort in solving.

Solving Methods Compared

Four main methods exist for solving quadratic equations. Each has a context where it is most efficient, and knowing when to apply each one is as important as knowing how to apply it.

Method 1
Factoring

When to use: When the discriminant is a perfect square and the coefficients are small integers. Factoring is the fastest method when it applies cleanly.

What it reveals: The factored form a(x − r1)(x − r2) = 0 exposes both roots at a glance. The Zero Product Property (if A × B = 0 then A = 0 or B = 0) converts a product into two simple linear equations instantly.

Limitations: Cannot be applied when roots are irrational or complex. Becomes guesswork when coefficients are large or no integer factor pair sums to b.

Worked example: x² − 5x + 6 = 0. Find two numbers that multiply to 6 and sum to −5: they are −2 and −3. So (x − 2)(x − 3) = 0, giving x = 2 or x = 3.

Conceptual note: Factoring transforms a sum of terms into a product of factors. Once you have a product equal to zero, each factor can be set to zero independently.

Method 2
Quadratic Formula

When to use: Always. The formula x = (−b ± √(b² − 4ac)) / 2a works for every quadratic, regardless of whether roots are rational, irrational, or complex.

What it reveals: The formula encodes the full structure: the discriminant predicts root type, −b / 2a is the axis of symmetry, and √Δ is the half-distance between the two roots along the x-axis.

Limitations: Requires careful arithmetic. Errors in computing Δ or forgetting the ± to produce both roots are the most common mistakes.

Worked example: 2x² − 7x + 3 = 0. a = 2, b = −7, c = 3. Δ = 49 − 24 = 25. x = (7 ± 5) / 4. x1 = 3, x2 = 1/2.

Conceptual note: The quadratic formula is the result of applying completing the square to the general form ax² + bx + c = 0. It is not a separate invention but the same algebraic technique carried through symbolically.

Method 3
Completing the Square

When to use: When converting to vertex form, deriving the quadratic formula, or when a = 1 and b is even, keeping the half-coefficient a clean integer.

What it reveals: Produces vertex form a(x − h)² + k directly, exposing the turning point (h, k) and the axis of symmetry. Makes the geometry of the parabola explicit.

Limitations: More steps than the formula for pure root-finding. Sign errors when adjusting the constant are common. Less efficient as a general-purpose solver.

Worked example: x² + 6x + 5 = 0. Move constant: x² + 6x = −5. Add (3)² = 9 to both sides: (x + 3)² = 4. Take roots: x + 3 = ±2. x = −1 or x = −5.

Conceptual note: Adding (b/2a)² to both sides does not change the equation's truth, only its form. The "completed square" is a perfect square trinomial that can be solved by a single square root.

Method 4
Graphing

When to use: When visualizing the roots, understanding function behavior, checking an analytical result, or estimating roots before algebraic work.

What it reveals: Shows the roots as x-intercepts, the vertex as the turning point, whether the parabola opens up or down, and whether the discriminant is positive, zero, or negative, all from a single visual.

Limitations: Gives only approximate roots unless they are integers. Graphical precision depends heavily on the scale and tools used.

Worked example: For y = x² − 4, the parabola clearly crosses the x-axis at x = −2 and x = 2, confirming two real roots without any formula computation.

Conceptual note: Graphing connects the algebraic notion of a root (a value making the equation zero) to the geometric notion of an x-intercept (a point on the curve at height zero). Both refer to the same mathematical object seen from two different perspectives.

Common Mistakes

These eight errors account for the vast majority of wrong answers in quadratic equation problems. Understanding why each mistake occurs is the first step toward reliably avoiding it.

1. Not setting the equation equal to zero before solving

What goes wrong: Applying the quadratic formula to x² + 3x = 10 without rearranging means c is treated as 0 instead of −10, producing entirely wrong roots.

Why it matters: The quadratic formula is derived for ax² + bx + c = 0 specifically. Any other arrangement represents a different equation with different solutions.

How to avoid it: Always move all terms to one side first. x² + 3x = 10 must become x² + 3x − 10 = 0 before identifying a, b, and c.

2. Misidentifying the signs of a, b, and c

What goes wrong: In 3x² − 2x + 5 = 0, reading b as +2 instead of −2, or c as −5 instead of +5, corrupts the discriminant and both roots.

Why it matters: The sign of each coefficient is part of its value. A sign error propagates through every calculation that follows, usually invisibly until verification fails.

How to avoid it: Write the full standard form explicitly and label a, b, and c with their signs before computing anything. In 3x² − 2x + 5 = 0: a = 3, b = −2, c = 5.

3. Dropping the negative sign in −b

What goes wrong: The formula uses −b in the numerator. When b = −4, then −b = +4, not −4. Forgetting to negate shifts both roots by a constant amount.

Why it matters: The axis of symmetry is −b / 2a. An error here shifts every root away from the correct value by exactly b / a.

How to avoid it: Write the −b substitution explicitly as an intermediate step. If b = −4, write −(−4) = +4 before proceeding. Never substitute and simplify simultaneously when signs are involved.

4. Miscalculating the discriminant

What goes wrong: (−5)² is 25, not −25. And 4ac must account for the signs of both a and c. Many errors arise when b is negative or when a and c have opposite signs.

Why it matters: The discriminant determines whether roots are real or complex and how many there are. A wrong Δ changes the entire nature and value of the solution.

How to avoid it: Compute b² first (it is always non-negative, since squaring removes the sign), then compute 4ac separately with full attention to signs, then subtract. Never combine these into a single step mentally.

5. Saying "no solution" when Δ < 0

What goes wrong: Concluding the equation is unsolvable when the discriminant is negative. Complex roots exist and are valid mathematical solutions.

Why it matters: In applied mathematics, complex roots carry physical meaning in circuit analysis, vibration theory, and control systems. Writing "no solution" is factually incorrect and can cause downstream errors.

How to avoid it: Use precise language: "no real solution." If the context requires complex roots, compute x = (−b ± i√|Δ|) / 2a and report both conjugate roots explicitly.

6. Dividing only part of the numerator by 2a

What goes wrong: Writing x = −b ± √Δ / 2a instead of (−b ± √Δ) / 2a. By standard order of operations, the division applies only to √Δ, leaving −b undivided.

Why it matters: This produces two completely wrong numerical answers. The entire expression −b ± √Δ is the numerator and must be divided by 2a as a single unit.

How to avoid it: Always write both roots as separate fractions with explicit parentheses: x1 = (−b + √Δ) / (2a) and x2 = (−b − √Δ) / (2a). This leaves no ambiguity about the scope of the division.

7. Reporting only one root when two exist

What goes wrong: Computing only the + case of ± and stopping, missing the second root entirely.

Why it matters: Quadratic equations have two roots (counting multiplicity). Reporting only one is an incomplete answer. In applied contexts, both roots may represent physically meaningful values, such as the two times when a projectile reaches a given height.

How to avoid it: Treat ± as two separate calculations every time. Write and simplify both x1 and x2 before deciding which roots are relevant to the problem context.

8. Trusting rounded decimals without verification

What goes wrong: After computing approximate roots like x ≈ 2.732 and x ≈ 0.268, not checking whether these values actually satisfy the original equation. Rounding errors can accumulate across several arithmetic steps.

Why it matters: A root that satisfies the equation only to two decimal places may introduce large errors in subsequent calculations. Verification catches these problems before they propagate.

How to avoid it: Substitute each root back into the original equation and confirm the result is zero (or acceptably close to zero for decimal approximations). Use exact surds or fractions wherever possible. This calculator performs automatic substitution verification and displays the results in the Verification panel.

Real-World Applications

Quadratic equations are not textbook abstractions. They arise wherever a quantity depends on the square of another, which occurs constantly in the physical, economic, and computational world.

Projectile motion

Under constant gravitational acceleration (ignoring air resistance), the height of any thrown or launched object follows h(t) = −(1/2)gt² + v0t + h0, a quadratic in time t. Here g is gravitational acceleration (9.8 m/s² near Earth’s surface), v0 is the initial vertical velocity, and h0 is the launch height. Setting h(t) = 0 and solving gives the exact times when the object hits the ground. The vertex gives the maximum height and the time at which it occurs. Every trajectory calculation in ballistics, sports science, and aerospace uses this quadratic structure.

Business revenue and profit optimization

When demand decreases linearly with price (a standard economic assumption), total revenue becomes price times quantity, the product of two linear functions in price, which is quadratic. The result is a downward-opening parabola. The vertex gives the price that maximizes revenue, answering the central business question of what to charge. Profit (revenue minus cost) is often also quadratic when fixed and variable costs are accounted for. Setting profit to zero and solving gives the two break-even prices, between which the business operates profitably.

Engineering design

Parabolic curves appear in arched bridges, suspension cable profiles, antenna dishes, solar concentrators, and telescope mirrors. The reflective property of a parabola (all parallel incoming rays reflect through the focal point) is used in every satellite dish and reflecting telescope. Engineers computing load distribution along a parabolic arch, cable tension in a suspension structure, or the focal length of a parabolic reflector all work with quadratic equations. The roots tell where the arch meets the supports; the vertex gives the peak height or minimum clearance.

Computer graphics and animation

Quadratic Bezier curves, defined by three control points, trace parabolic arcs and are the basis of vector graphics and font design. Collision detection in 2D games reduces to checking whether a quadratic in time (representing the distance squared between a moving object and a boundary) has real roots within a given interval. A positive discriminant means a collision occurs; a negative discriminant means the objects miss each other. Animation easing functions that produce smooth acceleration and deceleration are often quadratic in the time parameter.

Physics: motion under constant force

Every kinematics equation of the form s = ut + (1/2)at² is quadratic in time. Finding when a braking vehicle comes to rest, when a charged particle reaches a detector, or when two objects moving toward each other meet all require solving a quadratic in t. The discriminant predicts whether the scenario is physically possible: a negative discriminant means the object never reaches the target under the given conditions, which can reveal flawed assumptions in the physical setup.

Architecture and structural curves

Arched structures from doorways to large-span roofs use parabolic or approximately parabolic profiles because load distributes uniformly along a parabolic arch, making it mechanically efficient. Architects and structural engineers use quadratic functions to compute the height of an arch at any horizontal position, to ensure adequate clearance, and to calculate the total area under a curved surface. This shape appears from Roman aqueducts to modern reinforced concrete bridges for the same mathematical reason: the parabola minimizes material while distributing stress evenly.

Machine learning and optimization

The mean squared error (MSE) loss function used in linear regression is quadratic in the model’s weight parameters. Minimizing MSE requires finding where the derivative equals zero, which for a quadratic loss is equivalent to finding the vertex. This produces the closed-form normal equations for least-squares regression. Ridge regression adds a quadratic penalty term to the loss, and the entire problem remains quadratic, yielding an exact analytical solution. Understanding why quadratic losses are tractable analytically, while higher-degree losses generally are not, is a foundational insight in machine learning theory.

Economics and cost modeling

Total cost of production is often modeled as quadratic in output quantity when marginal cost rises with scale. The profit function (revenue minus cost) then becomes a downward-opening quadratic, with the vertex giving the profit-maximizing output level. Supply and demand models with quadratic supply or demand curves require solving a quadratic equation for the equilibrium price or quantity. Quadratic utility functions model investor risk preferences in portfolio optimization theory, where the mean-variance framework (maximizing expected return subject to a quadratic variance constraint) is solved by quadratic programming.

Advanced Notes

Vertex formula

For f(x) = ax² + bx + c, the vertex is at (h, k) where h = −b / (2a) and k = f(h) = c − b² / (4a). The vertex is the global minimum when a > 0 and the global maximum when a < 0. Every optimization problem involving a single quadratic function reduces to finding the vertex, which requires no calculus.

Axis of symmetry

The vertical line x = −b / (2a) is the axis of symmetry of the parabola. The two roots are always symmetric about this line: if x1 and x2 are the roots, then (x1 + x2) / 2 = −b / (2a). The axis of symmetry is also the x-coordinate of the vertex, lying exactly midway between the two roots. This symmetry means that if you know one root and the axis, you can find the other root without the formula: x2 = 2 × axis − x1.

Vieta’s formulas

Named after the French mathematician François Viète, these formulas relate the roots of ax² + bx + c = 0 directly to the coefficients without requiring the roots to be found: r1 + r2 = −b / a and r1 × r2 = c / a. These hold for all roots, real or complex. If your two computed roots fail either condition, at least one is wrong. Vieta’s formulas also explain the structure of the factored form: a(x − r1)(x − r2) expands to ax² − a(r1 + r2)x + a(r1 × r2), which must match ax² + bx + c exactly.

Repeated roots and tangency

When Δ = 0, the quadratic has a repeated root x = −b / (2a) with multiplicity 2. The polynomial factors as a(x − r)² = 0. Geometrically, the parabola is tangent to the x-axis at x = r: the curve touches the axis at that single point without crossing through it. Tangency is the borderline case between two crossings (Δ > 0) and no crossing (Δ < 0). In calculus terms, the curve has the same value and the same slope (zero) as the x-axis at the point of tangency.

Complex roots

When Δ < 0, write Δ = −|Δ|. The roots are x = (−b ± i√|Δ|) / (2a), where i² = −1. These roots are always complex conjugates. If one root is p + qi, the other is p − qi. Their sum is 2p = −b / a (real) and their product is p² + q² = c / a (also real), both consistent with Vieta’s formulas even in the complex case. The real part p = −b / (2a) is the x-coordinate of the axis of symmetry, and the imaginary part q = √|Δ| / (2|a|) measures how far the vertex is from the x-axis in a scaled sense.

Numerical stability in the quadratic formula

The standard formula can lose precision through catastrophic cancellation when b and √Δ are nearly equal in magnitude. For example, when b = 1000 and √Δ ≈ 999.999, computing −b + √Δ ≈ 0.001 as a difference of two large numbers amplifies rounding error enormously. A numerically stable approach: compute the root with the larger magnitude first as x1 = (−b − sign(b) × √Δ) / (2a), then use the Vieta product relation to get x2 = c / (a × x1). This avoids the dangerous subtraction and is the approach used in scientific computing libraries.

Why completing the square leads to the quadratic formula

Starting from ax² + bx + c = 0, divide by a: x² + (b/a)x + c/a = 0. Move c/a: x² + (b/a)x = −c/a. Add (b/2a)² to both sides: (x + b/2a)² = b²/(4a²) − c/a = (b² − 4ac) / (4a²). Take the square root: x + b/2a = ±√(b² − 4ac) / (2a). Subtract b/2a: x = (−b ± √(b² − 4ac)) / (2a). Every step is reversible and the derivation uses only basic algebra. The quadratic formula is the inevitable result of completing the square in general form, not a separate discovery.

Practice Problems

Work through each problem independently before revealing the answer. Twelve problems are arranged in three tiers: four easy (direct recognition and rearrangement), four intermediate (formula and completing the square), and four advanced (parameter analysis, complex roots, and applied word problems).

Easy

Easy · P1
Solve: x² − 9 = 0
Answer: x = 3 and x = −3

Explanation: Recognize the difference of squares: (x − 3)(x + 3) = 0. Alternatively rearrange to x² = 9 and take the square root directly. Δ = 0 − 4(1)(−9) = 36, confirming two distinct real roots.

Recommended method: Direct rearrangement or factoring as a difference of squares.
Easy · P2
Solve: x² + 6x + 9 = 0
Answer: x = −3 (repeated root)

Explanation: Recognize the perfect square trinomial: (x + 3)² = 0. The discriminant is Δ = 36 − 36 = 0, confirming a single repeated root. The parabola is tangent to the x-axis at x = −3 and does not cross through it.

Recommended method: Factoring as a perfect square.
Easy · P3
Solve: x² − 5x + 6 = 0
Answer: x = 2 and x = 3

Explanation: Factor by finding two numbers that multiply to 6 and sum to −5: they are −2 and −3. So (x − 2)(x − 3) = 0. Verify with Vieta’s formulas: sum = 2 + 3 = 5 = −b/a and product = 6 = c/a. Both conditions are satisfied.

Recommended method: Factoring.
Easy · P4
Solve: 2x² − 8 = 0
Answer: x = 2 and x = −2

Explanation: Divide both sides by 2: x² = 4. Take the square root: x = ±2. Because b = 0 in this equation, the axis of symmetry is x = 0 and the two roots are symmetric about the y-axis.

Recommended method: Direct rearrangement.

Intermediate

Intermediate · P5
Solve: 2x² − 7x + 3 = 0
Answer: x = 3 and x = 1/2

Explanation: a = 2, b = −7, c = 3. Δ = 49 − 24 = 25. x = (7 ± 5) / 4. x1 = 12/4 = 3, x2 = 2/4 = 1/2. Verification: 2(9) − 7(3) + 3 = 18 − 21 + 3 = 0. Correct for both roots.

Recommended method: Quadratic formula.
Intermediate · P6
Solve: x² + 4x − 12 = 0
Answer: x = 2 and x = −6

Explanation: Factor by finding two numbers that multiply to −12 and sum to 4: they are 6 and −2. So (x + 6)(x − 2) = 0. Check Vieta: sum = 2 + (−6) = −4 = −b/a and product = −12 = c/a. Both match.

Recommended method: Factoring.
Intermediate · P7
Solve: 3x² + 2x − 1 = 0
Answer: x = 1/3 and x = −1

Explanation: a = 3, b = 2, c = −1. Δ = 4 + 12 = 16. x = (−2 ± 4) / 6. x1 = 2/6 = 1/3, x2 = −6/6 = −1. Verification: 3(1/9) + 2(1/3) − 1 = 1/3 + 2/3 − 1 = 0. Correct.

Recommended method: Quadratic formula.
Intermediate · P8
Solve x² − 2x − 15 = 0 by completing the square
Answer: x = 5 and x = −3

Explanation: Move the constant: x² − 2x = 15. Add (−1)² = 1 to both sides: (x − 1)² = 16. Take the square root: x − 1 = ±4. So x = 5 or x = −3. Verification: 25 − 10 − 15 = 0 and 9 + 6 − 15 = 0. Both roots confirmed.

Recommended method: Completing the square (as required by the problem).

Advanced

Advanced · P9
Solve: x² + x + 1 = 0. State the roots and interpret the result geometrically.
Answer: No real solutions. Complex roots: x = (−1 ± i√3) / 2

Explanation: Δ = 1 − 4 = −3. Since Δ < 0, the formula gives x = (−1 ± i√3) / 2. Geometrically, the parabola y = x² + x + 1 has its vertex at (−1/2, 3/4). Because the vertex is above the x-axis and the parabola opens upward (a = 1 > 0), the entire curve is above y = 0 for all real x. It never crosses the x-axis, confirming no real roots.

Recommended method: Quadratic formula with discriminant analysis.
Advanced · P10
A ball is thrown upward at 20 m/s from a height of 5 m. Its height is h(t) = −5t² + 20t + 5. When does it hit the ground?
Answer: t = 2 + √5 ≈ 4.24 seconds

Explanation: Set h(t) = 0: −5t² + 20t + 5 = 0. Divide by −5: t² − 4t − 1 = 0. Δ = 16 + 4 = 20. t = (4 ± √20) / 2 = 2 ± √5. The negative root t = 2 − √5 ≈ −0.24 represents a time before launch and is discarded by the physical context. The ball hits the ground at t = 2 + √5 ≈ 4.24 seconds.

Recommended method: Quadratic formula with context filtering of the negative root.
Advanced · P11
The product of two consecutive integers is 182. Find all such pairs.
Answer: 13 and 14, or −14 and −13

Explanation: Let the integers be n and n + 1. Then n(n + 1) = 182, giving n² + n − 182 = 0. Δ = 1 + 728 = 729 = 27². n = (−1 ± 27) / 2. n = 13 (giving the pair 13, 14) or n = −14 (giving the pair −14, −13). Verify both: 13 × 14 = 182 and (−14) × (−13) = 182. Both pairs are valid solutions.

Recommended method: Setting up a word problem as a quadratic equation, then applying the formula.
Advanced · P12
For what values of k does kx² + 6x + 3 = 0 have two distinct real roots?
Answer: k < 3 and k ≠ 0

Explanation: For two distinct real roots, we need Δ > 0. Here Δ = 36 − 12k > 0, which gives 36 > 12k, so k < 3. Additionally, k ≠ 0 because when k = 0 the equation is no longer quadratic (it becomes 6x + 3 = 0, a linear equation with one root). When k = 3, Δ = 0 and the roots coincide. When k > 3, Δ < 0 and no real roots exist. The complete answer is k < 3 and k ≠ 0.

Recommended method: Discriminant analysis with a parameter.

Knowledge Check

Six questions testing conceptual understanding. Select an option to reveal the explanation. Questions cannot be retried after answering.

Q1. The discriminant of a quadratic equation is −9. What does this imply?
A negative discriminant means √Δ is imaginary. The quadratic formula produces complex conjugate roots of the form (−b ± i√9) / 2a. The equation is fully solvable algebraically; the solutions simply lie outside the real number line. The correct statement is "no real solution," not "no solution." Option d is always false since the formula works for every quadratic.
Q2. If the leading coefficient a is negative, what happens to the parabola?
The sign of a determines the direction the parabola opens. Positive a means opens upward with a minimum vertex. Negative a means opens downward with a maximum vertex. Width depends on the magnitude of |a|, not its sign. Whether real roots exist depends on the discriminant, not the sign of a alone: a negative-a parabola can still cross the x-axis twice if its vertex is below the axis.
Q3. The discriminant is exactly zero. How many x-intercepts does the graph have?
When Δ = 0, the ± term in the quadratic formula vanishes completely and the formula yields x = −b / (2a), a single unique value. The parabola is tangent to the x-axis at the vertex, touching it at exactly one point without crossing through it. Algebraically, the quadratic factors as a(x − r)² = 0, which has a repeated root at x = r with multiplicity 2.
Q4. Which solving method is most reliable when factoring clearly fails?
The quadratic formula x = (−b ± √(b² − 4ac)) / 2a is universally valid for any quadratic equation, producing exact rational, irrational, or complex roots as appropriate. Factoring is faster when it works, but it only works cleanly when the roots are rational and the coefficients are small. Graphical estimates are approximate. Dropping the x² term changes the problem entirely. The quadratic formula is always the safe and reliable fallback.
Q5. For x² − 6x + 9 = 0, what does Vieta’s formula predict for the sum of the roots?
Vieta’s formula gives root sum = −b / a. Here a = 1 and b = −6, so sum = −(−6) / 1 = 6. This equation has a repeated root x = 3 (since Δ = 36 − 36 = 0), and 3 + 3 = 6 confirms the formula. The product of the roots = c / a = 9, and indeed 3 × 3 = 9. Both Vieta conditions are satisfied.
Q6. A quadratic has roots x = 2 and x = −5. Which equation produced these roots?
Using Vieta’s formulas: sum = 2 + (−5) = −3 = −b / a, so b = 3; product = 2 × (−5) = −10 = c / a, so c = −10. With a = 1 the equation is x² + 3x − 10 = 0. Confirm by factoring: (x − 2)(x + 5) = x² + 5x − 2x − 10 = x² + 3x − 10. The other options either use the wrong sign for b or the wrong product.

Frequently Asked Questions

An equation is quadratic when the highest power of its variable is exactly 2, expressible in the standard form ax² + bx + c = 0 with a ≠ 0. The name comes from the Latin quadratus (squared). The critical requirement is that the x² term must be present and nonzero: if a = 0, the equation reduces to bx + c = 0, which is linear. The squared term is what creates the parabolic graph, the two-root structure, and the discriminant analysis. An equation like 4x² − 9 = 0 is quadratic (a = 4, b = 0, c = −9), while 4x − 9 = 0 is linear. An equation like x² + 3x = 10 is also quadratic but must first be rearranged to x² + 3x − 10 = 0 before the coefficients can be correctly identified and the quadratic formula applied.

The Fundamental Theorem of Algebra states that a polynomial of degree n has exactly n roots in the complex number system, counting multiplicity. A quadratic (degree 2) therefore has exactly two roots, though they may be real or complex. In the real number domain, the parabola y = ax² + bx + c is a smooth, continuous curve that opens in one direction and is symmetric about its axis. A smooth curve of this shape can cross any horizontal line, including y = 0, at most twice. The discriminant tells us whether both roots are real and distinct (Δ > 0), whether both roots coincide at one real point (Δ = 0), or whether both roots are complex with no real-valued solution (Δ < 0).

The discriminant Δ = b² − 4ac is best understood as a signed measure of the geometric gap between the parabola and the x-axis. The vertex height is k = c − b² / (4a), and the discriminant satisfies Δ = −4ak. When the vertex is below the x-axis and the parabola opens upward (or vice versa), the gap is negative and Δ is positive, predicting two crossings. When the vertex sits exactly on the axis, the gap is zero and Δ = 0, giving one tangent point. When the vertex is on the same side of the axis as the direction the parabola opens, there is no crossing and Δ < 0. Algebraically, Δ appears under the square root in the quadratic formula: a positive value yields two real roots, zero yields one, and a negative value makes the root imaginary. The discriminant therefore predicts root type using only addition, subtraction, multiplication, and squaring, without needing the full formula or a graph.

Factoring is faster and more illuminating when the quadratic has small integer coefficients and the discriminant is a perfect square (0, 1, 4, 9, 16, 25, and so on), because in these cases the roots are rational and discoverable by inspection. For example, x² − 7x + 12 = 0 factors immediately to (x − 3)(x − 4) = 0, with no calculation beyond recognizing that −3 and −4 multiply to 12 and add to −7. Factoring also reveals the relationship between roots and coefficients visually through the factored form. However, factoring becomes impractical when roots are irrational (involving square roots), when coefficients are large, or when the discriminant is not a perfect square. In those cases, the quadratic formula is not just an alternative but the reliable default. A useful heuristic: compute the discriminant first. If it is a perfect square, try factoring. If not, go straight to the formula.

Completing the square works because every quadratic expression can be rewritten as a perfect square trinomial plus a constant. Starting from x² + bx, notice that (x + b/2)² = x² + bx + (b/2)². This means x² + bx = (x + b/2)² − (b/2)². By adding and subtracting (b/2)², we rearrange the original equation without changing its truth. Once the left side is a perfect square, taking the square root of both sides is a single trivial step. The technique is not an algebraic trick but a systematic method for converting any quadratic into a form solvable by a single square root operation. It also provides a direct path from the general quadratic ax² + bx + c = 0 to the quadratic formula: applying completing the square symbolically to the general form produces x = (−b ± √(b² − 4ac)) / 2a step by step, proving the formula rather than just using it.

The vertex is the turning point of the parabola and the most informationally rich single point on the quadratic curve. Its x-coordinate h = −b / (2a) is the axis of symmetry, the line about which the two roots are mirror images. Its y-coordinate k = f(h) is the minimum value of the function when a > 0, or the maximum value when a < 0. In applied problems, the vertex answers optimization questions directly: the maximum height of a projectile, the price that maximizes revenue, or the output level that minimizes average cost. The vertex also lets you immediately assess whether real roots exist: if a > 0 and k > 0, the parabola never crosses the x-axis (Δ < 0); if k = 0, it is tangent (Δ = 0); if k < 0, it crosses twice (Δ > 0). Knowing the vertex is equivalent to having the equation in vertex form a(x − h)² + k = 0, which is solved by rearranging to (x − h)² = −k/a and taking the square root.

A quadratic has no real roots when the parabola does not intersect the x-axis at any real point. This occurs when the vertex is on the same side of the x-axis as the direction the parabola opens: for example, when a > 0 (opens upward) and the vertex is above the axis, the entire parabola remains above y = 0. Algebraically, Δ = b² − 4ac < 0, so the quadratic formula requires the square root of a negative number, producing an imaginary value. The roots do exist as complex conjugates: x = (−b ± i√|Δ|) / 2a. They always appear as a conjugate pair with real part −b / (2a) (the axis of symmetry) and imaginary part ±√|Δ| / (2|a|). In applied problems, a negative discriminant signals that the scenario described is impossible under the given conditions: for instance, a projectile launched from ground level cannot reach a target altitude if it was not thrown hard enough, or a company cannot achieve a specific revenue target at any price within the model assumptions.

The primary check is substitution: take each root r and substitute it back into the original equation. If ar² + br + c equals zero (or is sufficiently close to zero for decimal approximations), the root is confirmed. This works regardless of the solving method used. A complementary check applies Vieta’s formulas: the sum of your two roots must equal −b / a, and the product must equal c / a. If both conditions hold simultaneously, the roots are almost certainly correct. A third consistency check involves the discriminant: if you found two distinct real roots, confirm Δ > 0; if one repeated root, Δ should be exactly 0; if complex roots, Δ should be negative. Any mismatch between your results and the discriminant indicates an error somewhere. This calculator performs automatic substitution verification for every solution and displays the result in the Verification panel, removing the need to check by hand.