Quadratic Equation Solver
Solve ax² + bx + c = 0: the discriminant, real or complex roots, and the vertex of the parabola.
Equation
x² − 5x + 6 = 0
Two real roots
x₁ = 2,
- Discriminant
- Δ = 1
- Vertex
- (2.5, −0.25)
- Axis of symmetry
- x = 2.5
Because a > 0, the parabola opens upward and the vertex is its lowest point.
Frequently Asked Questions
The solutions of ax² + bx + c = 0 (with a ≠ 0) are x = (−b ± √(b² − 4ac)) / 2a. For x² − 5x + 6 = 0 this gives x = (5 ± √1) / 2, so x = 2 or x = 3.
The discriminant Δ = b² − 4ac decides the kind of roots. If Δ > 0 there are two different real roots, if Δ = 0 there is one repeated real root, and if Δ < 0 there are no real roots but two complex roots.
When Δ < 0 the formula needs the square root of a negative number. Using i = √−1, the roots are written p ± qi, where p = −b / 2a and q = √(−Δ) / 2|a|. For x² + 2x + 5 = 0 the roots are −1 ± 2i.
The vertex of y = ax² + bx + c lies at x = −b / 2a; put that x into the equation to get its y-value. The vertical line through the vertex, x = −b / 2a, is the axis of symmetry.
Then there is no x² term and the equation is linear: bx + c = 0 gives x = −c / b. If b is also 0, either every x is a solution (when c = 0) or there is no solution (when c ≠ 0).