Quadratic Equation Solver
Enter the coefficients a, b and c of ax² + bx + c = 0. Real roots are shown as plain decimals with up to 10 significant digits, complex roots in the form p + qi.
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Formula
- Discriminant:
D = b² − 4ac - Roots:
x = (−b ± √D) ÷ 2a. If D < 0 the roots are complex:−b ÷ 2a ± (√−D ÷ 2|a|) i - Vertex:
x = −b ÷ 2a,y = c − b² ÷ 4a
Worked example
x² − 5x + 6 = 0: D = 25 − 24 = 1, roots (5 ± 1) ÷ 2 = 3 and 2, vertex at x = 2.5, y = 6 − 6.25 = −0.25.
Frequently asked questions
How do you solve a quadratic equation?
Use the quadratic formula x = (-b ± √(b² − 4ac)) ÷ 2a. For x² − 5x + 6 = 0 the discriminant is 25 − 24 = 1, so x = (5 ± 1) ÷ 2, giving 3 and 2.
What does the discriminant tell me?
If b² − 4ac is positive there are two real roots, if zero one repeated root, and if negative two complex roots. For x² + 2x + 5 it is -16, so the roots are -1 ± 2i.
How do I find the vertex of a parabola?
The vertex is at x = -b ÷ 2a, and y is the equation's value there. For x² − 5x + 6 it is x = 2.5 and y = -0.25.
What if a is zero?
Then the equation is linear, not quadratic, and the formula would divide by zero. Solve bx + c = 0 instead, giving x = -c ÷ b.