Quadratic Equation Solver — Roots, Discriminant & Vertex
A quadratic equation has the form ax² + bx + c = 0, where a ≠ 0. Its solutions come from the quadratic formula: x = (−b ± √(b² − 4ac)) ÷ 2a. The expression under the square root, Δ = b² − 4ac, is the discriminant, and it determines the nature of the roots before you solve anything: two distinct real roots if Δ > 0, one repeated real root if Δ = 0, and two complex conjugate roots if Δ < 0. Enter the coefficients a, b, and c to get the discriminant, the roots (real or complex), and a full step-by-step solution.
Beyond the roots, the calculator reports the vertex of the parabola y = ax² + bx + c, located at x = −b ÷ 2a and y = −Δ ÷ 4a, which is the minimum point when a > 0 and the maximum when a < 0. The plotted graph shows where the parabola crosses the x-axis — exactly at the real roots — making the connection between algebra and geometry visible. This is the same reasoning used in physics for projectile motion and in economics for profit maximization.
How it works
x = (−b ± √Δ) ÷ 2a, with discriminant Δ = b² − 4ac. Vertex: xᵥ = −b ÷ 2a, yᵥ = −Δ ÷ 4a. Sum of roots: x₁ + x₂ = −b ÷ a; product: x₁ × x₂ = c ÷ a (Vieta's formulas).
Use cases
- Solving algebra homework with a verifiable step-by-step solution
- Checking whether an equation has real or complex roots via the discriminant
- Finding the vertex to determine the maximum or minimum of a quadratic function
- Computing when a projectile reaches the ground in physics problems
- Factoring a quadratic by first finding its roots
- Visualizing how changing a, b, and c reshapes the parabola
Frequently asked questions
How do you solve a quadratic equation with the quadratic formula?
Identify the coefficients a, b, and c in ax² + bx + c = 0, compute the discriminant Δ = b² − 4ac, then apply x = (−b ± √Δ) ÷ 2a. For x² − 5x + 6 = 0: Δ = 25 − 24 = 1, so x = (5 ± 1) ÷ 2, giving x₁ = 3 and x₂ = 2. The ± sign produces the two roots.
What does the discriminant tell you about the roots?
The discriminant Δ = b² − 4ac classifies the solutions without solving the equation. If Δ > 0 there are two distinct real roots and the parabola crosses the x-axis twice; if Δ = 0 there is one repeated real root and the parabola touches the axis at its vertex; if Δ < 0 there are no real roots — the solutions are two complex conjugates and the parabola never touches the x-axis.
What happens when the discriminant is negative?
The equation has no real solutions, but it still has two complex roots of the form x = (−b ± i√|Δ|) ÷ 2a, where i is the imaginary unit. For example, x² + 2x + 5 = 0 has Δ = 4 − 20 = −16, giving x = −1 ± 2i. Graphically, the parabola lies entirely above or below the x-axis.
How do you find the vertex of a parabola from the equation?
The vertex of y = ax² + bx + c is at xᵥ = −b ÷ 2a and yᵥ = −Δ ÷ 4a (equivalently, substitute xᵥ back into the function). When a > 0 the parabola opens upward and the vertex is the minimum; when a < 0 it opens downward and the vertex is the maximum. This point is what optimization problems ask for, such as the price that maximizes revenue.
Can a quadratic equation be solved without the quadratic formula?
Often, yes. If the quadratic factors nicely, you can find two numbers whose sum is −b/a and whose product is c/a (Vieta's formulas) — for x² − 5x + 6, those are 2 and 3. Completing the square also works and is how the quadratic formula is derived. Incomplete quadratics are even simpler: ax² + bx = 0 factors as x(ax + b) = 0, and ax² + c = 0 solves by isolating x².