Quadratic Equation Solver
Solve ax² + bx + c = 0 instantly — real or complex roots, discriminant, vertex and factored form, with the quadratic formula steps explained.
About the Quadratic Equation Solver
This quadratic equation solver finds the roots of ax² + bx + c = 0 using the quadratic formula x = (−b ± √(b² − 4ac)) ÷ 2a. Enter the three coefficients and it instantly returns the roots — real or complex — along with the discriminant, the vertex of the parabola and, when the roots are rational, the factored form of the expression.
The discriminant b² − 4ac tells the story before you solve: positive means two distinct real roots, zero means one repeated root, negative means a pair of complex conjugate roots. The solver labels each case and shows complex answers in standard p ± qi form rather than an error.
A step summary rewrites the working with your actual numbers, which makes the tool double as a checking aid for algebra homework, physics projectile problems and engineering calculations. Decimal and fractional coefficients are both accepted.
How to Use the Quadratic Equation Solver
- 1Enter coefficient a (it cannot be zero — that would make the equation linear).
- 2Enter coefficients b and c; use negative signs where needed.
- 3Read the roots, discriminant and vertex instantly.
- 4Check the factored form (when available) and the step-by-step summary.
Frequently Asked Questions
What is the quadratic formula?
For ax² + bx + c = 0 with a ≠ 0, the roots are x = (−b ± √(b² − 4ac)) ÷ 2a. The expression under the square root, b² − 4ac, is the discriminant. For x² − 5x + 6 = 0: x = (5 ± √1) ÷ 2, giving x = 3 and x = 2. The formula works for every quadratic, even when factoring is impossible.
What does the discriminant tell you?
The discriminant D = b² − 4ac determines the nature of the roots without solving. D > 0 gives two distinct real roots (the parabola crosses the x-axis twice); D = 0 gives one repeated real root (the vertex touches the axis); D < 0 gives two complex conjugate roots (the parabola never meets the x-axis). It also flags factorability: a perfect-square D means rational roots.
What are complex roots and why do they appear?
When the discriminant is negative, its square root is imaginary, so the roots take the form p ± qi where i = √(−1). Geometrically the parabola never touches the x-axis. Complex roots always arrive as a conjugate pair when coefficients are real. This solver formats them properly — for x² + 2x + 5 = 0 it returns x = −1 ± 2i.
How do I find the vertex of a parabola from the equation?
The vertex sits at x = −b ÷ 2a; substitute back to get the y-coordinate, which simplifies to c − b² ÷ 4a. For y = 2x² − 8x + 3, the vertex is at x = 2, y = −5. The vertex is the minimum point when a > 0 and the maximum when a < 0 — essential for optimisation and projectile-motion problems.
When does the solver show a factored form?
When a, b and c are integers and the discriminant is a perfect square, the roots are rational and the quadratic factors cleanly, so the solver displays it as a(x − r₁)(x − r₂) with the roots shown as exact fractions. If the discriminant is not a perfect square, the roots are irrational or complex and no clean factorization over the rationals exists.