Quadratic Equation Solver

Enter a, b, c and get the roots, discriminant, vertex, axis and full working.

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A quadratic equation solver that takes the three coefficients of ax² + bx + c = 0 and returns everything you need: the two roots (real or complex), the discriminant, the vertex and axis of symmetry of the parabola, and a numbered, copyable derivation that shows exactly how the answer was reached. It is built for students checking homework, teachers preparing worked examples, and anyone who needs a fast, reliable solution without installing software or sending data anywhere.

How it works

A quadratic is any equation that can be written in the standard form ax² + bx + c = 0, where a is not zero. The solver applies the quadratic formula:

x = (−b ± √(b² − 4ac)) ⁄ 2a

The quantity under the square root, b² − 4ac, is the discriminant and it controls the whole result. When the discriminant is positive the square root is a real number and the ± produces two distinct real roots. When it is exactly zero the ± collapses to a single value, giving one repeated real root that sits right on the vertex. When the discriminant is negative the square root is imaginary, so the solver writes the two roots as a complex conjugate pair p ± qi.

Alongside the roots, the tool computes the axis of symmetry at x = −b ⁄ 2a and substitutes that value back into the function to find the vertex — the turning point of the parabola. If you set a to zero the equation degenerates to a linear one, bx + c = 0, and the solver handles that case too, reporting the single root x = −c ⁄ b (or flagging an identity or no-solution when b is also zero).

Worked example

Take x² − 3x + 2 = 0, so a = 1, b = −3, c = 2. The discriminant is Δ = (−3)² − 4(1)(2) = 9 − 8 = 1, which is positive, so expect two real roots. The square root of 1 is 1, and the formula gives x = (3 ± 1) ⁄ 2, that is x = 2 and x = 1. The axis of symmetry is x = 3 ⁄ 2 = 1.5, and the vertex is (1.5, −0.25) — the lowest point of an upward-opening parabola. You can verify the roots by factoring: (x − 1)(x − 2) = 0.

Now try x² + 1 = 0, where a = 1, b = 0, c = 1. The discriminant is 0 − 4 = −4, which is negative, so the roots are complex: x = ±i. The solver reports them as 0 + 1i and 0 − 1i, a perfect conjugate pair.

EquationDiscriminantRoots
x² − 3x + 2 = 012 and 1
x² − 4x + 4 = 002 (double root)
x² + 1 = 0−40 ± 1i
2x² + 4x − 6 = 0641 and −3

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