Quadratic Equation Solver
Solve any quadratic equation and see not just the roots but the shape they come from. Enter the three coefficients and you get the discriminant with an explanation of what its sign means, the real or complex roots, the vertex, the axis of symmetry, the y-intercept, the factored form, and a drawn parabola showing where the curve meets the x-axis. The roots are computed in a way that avoids the precision loss the textbook formula suffers when one root is much smaller than the other: subtracting two nearly equal numbers destroys most of the significant digits, so the well-conditioned root is found first and the other derived from the product of the roots instead. The working is shown with your numbers substituted in.
How it works
Enter the three coefficients of ax² + bx + c = 0. Use 0 for any term that is missing — for x² = 4, that is a = 1, b = 0, c = −4.
As well as the roots you get the discriminant, the vertex, the axis of symmetry and the factored form, plus a graph showing where the parabola meets the x-axis.
Discriminant
Positive — the parabola crosses the x-axis at two distinct points, so there are two real roots.
The parabola
● roots · ● vertex
Factored form
The equation rewritten as a product of its roots.
(x - 1)(x - 2) = 0
Working
The quadratic formula with your numbers substituted in.
x = (−b ± √(b² − 4ac)) ÷ 2a
x = (−(-3) ± √((-3)² − 4 × 1 × 2)) ÷ (2 × 1)
x = (3 ± √1) ÷ 2
Frequently asked questions
What does the discriminant tell me?
The sign of b² − 4ac decides everything: positive means two distinct real roots, zero means one repeated root where the parabola touches the axis, and negative means the roots are a pair of complex conjugates.
How do I enter an equation with a missing term?
Use 0 for that coefficient. For x² = 4, rearrange to x² − 4 = 0 and enter a = 1, b = 0, c = −4.
Can it handle complex roots?
Yes. When the discriminant is negative both conjugate roots are given in a ± bi form rather than the solver simply reporting no solution.
What is the vertex used for?
It is the turning point — the maximum when the parabola opens downwards and the minimum when it opens upwards — which is what you want for optimisation and projectile problems.
Why do the roots not always factor neatly?
The factored form is only exact when the roots are rational. When they are irrational the roots are still given precisely, but the factorisation shown is a decimal approximation.
What happens if I enter a = 0?
Then it is a linear equation, not a quadratic. The solver says so and gives the single root rather than dividing by zero.
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