Algebra • Core Pillar

Quadratic Formula Calculator

The universal quadratic algebra calculator for evaluating real and complex roots of $ax^2 + bx + c = 0$, analyzing the discriminant $\Delta$, finding vertex coordinates $(h, k)$, and plotting dynamic parabolic curves.

Reviewed by Sanjay Samanta & Academic Review Board

Last Updated: September 2026

Real & Complex Roots Evaluation

Equation Coefficients

ax² + bx + c = 0

Coefficient a (Quadratic, a ≠ 0)

Coefficient b (Linear)

Coefficient c (Constant)

Example Equations:

Quadratic Solutions & Roots

Two Real Roots (Δ > 0)

Calculated Roots (x)

x₁ = 3, x₂ = 2

x² − 5x + 6 = 0

Discriminant (Δ) 1

Vertex (h, k) (2.5, -0.25)

Axis of Symmetry x = 2.5

Y-Intercept (0, 6)

Interactive Parabola Curve Plot

Parabola Roots Vertex

Step-by-Step Quadratic Formula Derivation

Direct Answer & Overview

Verified Educational Guide

How to Solve Quadratic Equations Using the Quadratic Formula

To solve any quadratic equation ax² + bx + c = 0, identify the coefficients a, b, and c (where a ≠ 0) and substitute them into the quadratic formula x = (−b ± √(b² − 4ac)) / (2a). The term under the square root (the discriminant Δ = b² − 4ac) indicates whether the solutions are two real numbers, one repeated real number, or two complex conjugates.

Primary Mathematical Formula Standard Mathematical Model

Standard Equation

ƒ(x)

Q.E.D.

x = rac{-b pm sqrt{b^2 - 4ac}}{2a} quad | quad ext{Discriminant } Delta = b^2 - 4ac

Evaluated with exact mathematical formulation • Rigorously verified

Exact Formula

Input Parameters

Required

Coefficient a (Quadratic term, a ≠ 0)

Coefficient b (Linear term)

Coefficient c (Constant term)

Expected Outputs

Calculated

Exact and Numerical Roots (x₁ and x₂)

Discriminant Value Δ and Nature of Roots Classification

Parabola Vertex Coordinates (h, k) and Axis of Symmetry

Interactive SVG Parabola Graph & Step-by-Step Derivation

Worked Numerical Example

Instant Verification

Solve x² − 5x + 6 = 0

→ a = 1, b = -5, c = 6. Δ = (-5)² - 4(1)(6) = 25 - 24 = 1. x = [5 ± √1] / 2 = [5 ± 1] / 2

x₁ = 3, x₂ = 2

Anatomy of a Quadratic Equation (ax² + bx + c = 0)

A quadratic equation is a second-order polynomial equation in a single variable x where the highest exponent is 2. In standard form, it contains three distinct components:

Quadratic Term (ax²)

a ≠ 0

Defines the curvature and opening direction of the parabola. If a > 0, the curve opens upward; if a < 0, it opens downward.

Linear Term (bx)

b ∈ ℝ

Determines the horizontal and vertical shift of the parabolic axis of symmetry off the y-axis.

Constant Term (c)

c ∈ ℝ

The y-intercept of the parabola, corresponding to the point (0, c) where the curve intersects the vertical axis.

The Quadratic Formula & Discriminant Analysis (Δ = b² − 4ac)

The term under the radical in the quadratic formula is called the discriminant (Δ). It reveals the algebraic nature and geometric count of intersections without calculating the exact roots:

Case 1: Δ > 0 Positive

2 Real Distinct Roots

The parabola intersects the horizontal x-axis at two distinct real coordinate points (x₁ ≠ x₂).

Case 2: Δ = 0 Zero

1 Real Repeated Root

The parabola is tangent to the x-axis; its vertex touches the horizontal axis exactly once (x = −b / 2a).

Case 3: Δ < 0 Negative

2 Complex Conjugate Roots

The parabola never crosses the x-axis. Roots exist in the complex plane as conjugate pairs (p ± qi).

Deriving the Formula via Completing the Square

The quadratic formula is derived directly from the standard quadratic equation using the algebraic technique of completing the square:

1. Start with standard form: ax² + bx + c = 0

2. Divide all terms by a: x² + (b/a)x + c/a = 0

3. Move the constant term to the right: x² + (b/a)x = −c/a

4. Add (b / 2a)² = b² / (4a²) to both sides to complete the perfect square trinomial:

x² + (b/a)x + b²/(4a²) = b²/(4a²) − c/a = (b² − 4ac) / (4a²)

5. Factor the left side as a binomial square: (x + b/(2a))² = (b² − 4ac) / (4a²)

6. Take the square root of both sides: x + b/(2a) = ±√(b² − 4ac) / (2a)

7. Isolate x: x = (−b ± √(b² − 4ac)) / (2a)

Parabola Geometry: Vertex (h, k), Axis of Symmetry & Extrema

Graphically, every quadratic function f(x) = ax² + bx + c traces a symmetrical U-shaped curve known as a parabola:

Vertex Coordinates (h, k)

The peak turning point of the curve is located at:

h = −b / (2a), k = c − (b² / 4a)

Axis of Symmetry & Vertex Form

The vertical line of bilateral mirror symmetry is x = h. Vertex form is:

y = a(x − h)² + k

Real-World Applications (Physics Trajectories & Optimization)

Quadratic equations model physical laws and optimization systems across multiple scientific fields:

Kinematics & Projectile Trajectories

Under uniform gravity g, the vertical position of an object is h(t) = −½gt² + v₀t + h₀. Solving for h(t) = 0 gives the exact landing time.

Revenue & Profit Maximization

Price-demand curves create parabolic revenue functions R(p) = −ap² + bp. The vertex gives the optimal price that maximizes profit.

Structural Engineering & Suspension Bridges

Suspension bridge main cables and parabolic arch bridges distribute compressive gravitational load uniformly to anchor abutments.

Optics & Satellite Reflectors

Paraboloid surfaces reflect incoming parallel light or radio waves directly to a single focus point, powering satellite dishes and telescope mirrors.

Step-by-Step Worked Examples

Example 1: Solving 2x² − 8x + 6 = 0 Two Real Roots

1. Identify coefficients: a = 2, b = −8, c = 6.

2. Calculate discriminant: Δ = (−8)² − 4(2)(6) = 64 − 48 = 16 (> 0 ⟹ 2 real roots).

3. Apply formula: x = [−(−8) ± √16] / (2 × 2) = [8 ± 4] / 4.

4. Evaluate roots: x₁ = (8 + 4)/4 = 3, x₂ = (8 − 4)/4 = 1.

Example 2: Solving x² + 2x + 5 = 0 Complex Conjugate Roots

1. Coefficients: a = 1, b = 2, c = 5.

2. Discriminant: Δ = (2)² − 4(1)(5) = 4 − 20 = −16 (< 0 ⟹ Complex roots).

3. Square root of negative discriminant: √−16 = 4i.

4. Evaluate: x = [−2 ± 4i] / 2 = −1 ± 2i.

Common Calculation Pitfalls & Sign Errors

Sign Error on −b

If b = −5, then −b = −(−5) = +5. Dropping the double negative is the #1 algebraic error students make.

Fractional Bar Division Scope

The denominator 2a divides the ENTIRE numerator [−b ± √Δ], not just the square root term.

Squaring Negative b Values

(−4)² = +16, never −16. The first term of the discriminant b² is always positive or zero.

Fact-Checked & Verified • Computational Accuracy Standards

Updated July 2026 • Editorial Policy

Authored By

Sanjay Samanta

Lead Developer & Founder of Basic Math Tools. Specializes in browser-native computational algorithms and applied mathematics.

Reviewed & Verified By

Academic Review Board

Mathematics & curriculum specialists. Audited against standard algebraic and arithmetic principles.

Found an error or have an improvement suggestion? Report a calculation issue

Frequently Asked Questions

What is the quadratic formula and when is it used?

The quadratic formula is x = (−b ± √(b² − 4ac)) / (2a). It provides the exact analytical solutions (roots) for any second-degree polynomial equation in standard form: ax² + bx + c = 0, where a ≠ 0. It works universally for all quadratic equations, including those that cannot be factored using simple rational integers.

What does the discriminant (b² − 4ac) tell you about the roots?

The discriminant Δ = b² − 4ac determines the number and nature of the solutions: (1) If Δ > 0, the equation has two distinct real roots (parabola crosses the x-axis twice); (2) If Δ = 0, the equation has exactly one real repeated root (parabola vertex touches the x-axis); (3) If Δ < 0, the equation has two complex conjugate roots in the form p ± qi (parabola never crosses the x-axis).

How do you find the vertex (h, k) and axis of symmetry of a parabola?

The x-coordinate of the vertex (and the equation of the axis of symmetry) is h = −b / (2a). The y-coordinate is found by substituting h back into the quadratic function: k = f(h) = c − b² / (4a). The vertex represents the absolute minimum point if a > 0 (opens upward) or absolute maximum point if a < 0 (opens downward).

How do you convert a standard quadratic equation to vertex form?

A standard quadratic equation y = ax² + bx + c is converted to vertex form y = a(x − h)² + k by completing the square or computing the vertex coordinates directly: h = −b / (2a) and k = c − ah². For example, y = x² − 6x + 5 has h = 3 and k = −4, giving vertex form y = (x − 3)² − 4.

Can the quadratic coefficient "a" ever equal zero?

No. If a = 0, the x² term vanishes and the equation reduces to a first-degree linear equation (bx + c = 0), which is solved by simple division (x = −c / b) and cannot be evaluated with the quadratic formula because division by 2a would result in division by zero.

How does the quadratic formula handle complex numbers with negative discriminants?

When Δ < 0, the square root of the negative discriminant produces the imaginary unit i (where i = √−1). The solutions are expressed as a complex conjugate pair: x = −b/(2a) ± (√(−Δ)/(2a)) · i.

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