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Quadratic Equation Solver

Solve quadratic equations (ax² + bx + c = 0) with real and complex roots. Free online solver. No signup, 100% private, browser-based.

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Quadratic Equation Solver

Solve ax^2 + bx + c = 0.

Root 1

Complex

Root 2

Complex

How it works

A quadratic equation has the form ax² + bx + c = 0, where a ≠ 0. Every quadratic equation has exactly two solutions in the complex numbers (counting multiplicity) — a consequence of the Fundamental Theorem of Algebra. The Quadratic Equation Solver computes both roots, shows the discriminant, and identifies whether roots are real-distinct, real-equal, or complex conjugate pairs.

**The Quadratic Formula** x = (−b ± √(b² − 4ac)) / (2a)

The derivation is completing the square: add (b/2a)² to both sides to create a perfect square, then take the square root. This formula works for all real and complex coefficients.

**The Discriminant (Δ = b² − 4ac)** - Δ > 0: two distinct real roots (parabola crosses x-axis twice) - Δ = 0: one repeated real root (parabola touches x-axis exactly once — vertex on x-axis) - Δ < 0: two complex conjugate roots (parabola doesn't cross x-axis)

**Vieta's Formulas** Sum of roots: x₁ + x₂ = −b/a. Product of roots: x₁ × x₂ = c/a. These let you verify solutions quickly: if you solve ax² + bx + c = 0 and get x₁, x₂, check that they sum to −b/a and multiply to c/a.

**Numerical precision note** For coefficients with very different magnitudes (e.g., a = 1, b = 10⁸, c = 1), the standard formula loses precision through catastrophic cancellation. The numerically stable variant uses −2c / (−b ∓ √Δ) for one root. The solver uses the stable formula automatically when b² >> 4ac.

**Real-world applications** Projectile motion (find when height = 0), profit optimisation (find break-even quantity), geometric area problems (find side lengths from area constraints), financial modelling (find interest rate from payment equation).

Privacy: all computation runs in the browser. No data is transmitted.

Frequently Asked Questions

What does it mean when the discriminant is negative?
When b² − 4ac < 0, the quadratic has two complex conjugate roots: x = (−b ± i√|Δ|) / (2a), where i = √−1. Geometrically, the parabola y = ax² + bx + c doesn't intersect the x-axis at all — it floats entirely above (if a > 0) or below (if a < 0) the x-axis. Complex roots still satisfy the equation — they just require the complex number system. In engineering, complex roots of characteristic equations indicate oscillatory (underdamped) behaviour in differential equations.
How do I verify my quadratic solutions are correct?
Two quick verification methods: (1) Substitute each root back into the original equation and check it equals zero. (2) Use Vieta's formulas: sum of roots = −b/a; product of roots = c/a. If x₁ = 2 and x₂ = 3 for the equation x² − 5x + 6 = 0: sum = 2 + 3 = 5 = −(−5)/1 ✓; product = 2 × 3 = 6 = 6/1 ✓. Vieta's check is fast and catches most algebra errors without substituting complex expressions.
Can I solve a quadratic by factoring instead of using the formula?
Yes — factoring is faster when the roots are integers or simple fractions. x² − 5x + 6 = 0: find two numbers that multiply to 6 and add to −5: −2 and −3. Factor: (x − 2)(x − 3) = 0 → x = 2 or x = 3. Factoring works when roots are rational; if the discriminant is not a perfect square (e.g., Δ = 7), the roots are irrational and factoring over integers is impossible — use the quadratic formula directly.
What are real-world problems that lead to quadratic equations?
Projectile motion: if a ball is thrown upward at 20 m/s from height 1.5 m, when does it hit the ground? h(t) = −4.9t² + 20t + 1.5 = 0 → solve with quadratic formula. Area problems: a rectangular field has area 600 m² and one side is 5 m longer than the other → x(x+5) = 600 → x² + 5x − 600 = 0. Break-even analysis: profit = −2q² + 100q − 500 = 0 → find break-even quantities. Optics: lens equation manipulation; electrical engineering: RLC circuit resonance.