Quadratic Equation Solver
Solve any quadratic equation ax² + bx + c = 0 — get the roots, discriminant and vertex.
Real, repeated or complex roots are all handled automatically.
What Is a Quadratic Equation Solver?
A Quadratic Equation Solver finds the values of x that satisfy any equation of the form ax² + bx + c = 0, where a is not zero. These solutions are called the roots of the equation, and they are the points where the matching parabola crosses the x-axis. Enter the three coefficients a, b and c, and the calculator returns the roots, the discriminant and the vertex of the curve.
Quadratics appear throughout algebra, physics and engineering — modelling projectile paths, areas, profit curves and more. Solving them by hand means applying the quadratic formula carefully and checking the discriminant, which is easy to get wrong with signs. This tool does the steps reliably and also tells you whether the roots are two distinct real numbers, one repeated value, or a pair of complex numbers.
The Quadratic Formula and Discriminant
The roots come from the quadratic formula:
x = (−b ± √D) ÷ 2a, where D = b² − 4ac is the discriminant.
The discriminant D decides the nature of the roots:
- D > 0 — two distinct real roots; the parabola cuts the x-axis at two points.
- D = 0 — one repeated real root; the parabola just touches the x-axis.
- D < 0 — two complex (conjugate) roots; the parabola never meets the x-axis.
The turning point of the parabola, its vertex, sits at x = −b ÷ 2a, exactly halfway between the two roots when they are real. Substituting that x back into the equation gives the vertex y-value, the minimum of the curve when a is positive or the maximum when a is negative.
Worked Example
Solve x² − 3x + 2 = 0, so a = 1, b = −3 and c = 2.
Step 1 – Discriminant: D = b² − 4ac = (−3)² − 4(1)(2) = 9 − 8 = 1. Since D is positive, expect two distinct real roots.
Step 2 – Apply the formula: x = (−(−3) ± √1) ÷ (2 × 1) = (3 ± 1) ÷ 2.
This gives x = (3 + 1) ÷ 2 = 2 and x = (3 − 1) ÷ 2 = 1.
Step 3 – Vertex: x = −b ÷ 2a = 3 ÷ 2 = 1.5, which lies exactly between the roots 1 and 2. So the equation has roots x = 1 and x = 2, and the parabola dips to its lowest point at x = 1.5.
Factoring and Checking Your Roots
Not every quadratic needs the full formula. When the numbers are friendly, factoring is faster: x² − 3x + 2 factors as (x − 1)(x − 2) = 0, giving roots 1 and 2 directly. The formula is the safe fallback that always works, even when factoring is awkward or the roots are irrational or complex.
You can verify any pair of roots with two quick checks based on the coefficients:
- Sum of roots = −b ÷ a. Here 1 + 2 = 3 = −(−3) ÷ 1.
- Product of roots = c ÷ a. Here 1 × 2 = 2 = 2 ÷ 1.
If your roots satisfy both relationships, they are correct. When D is negative the roots become a complex conjugate pair such as p + qi and p − qi, and these same sum and product rules still hold, which is a handy way to confirm complex answers.
Frequently Asked Questions
The quadratic formula solves ax² + bx + c = 0 and is x = (−b ± √(b² − 4ac)) ÷ 2a. The plus and minus give the two roots, and the expression under the root sign, b² − 4ac, is the discriminant.
The discriminant D = b² − 4ac reveals the type of roots without solving fully. If D is greater than zero there are two real roots, if D equals zero there is one repeated root, and if D is less than zero the roots are complex.
When the discriminant is negative, the square root involves the imaginary unit i, where i² = −1. The roots then form a conjugate pair such as 2 + 3i and 2 − 3i, and the parabola does not cross the x-axis at any real point.
The x-coordinate of the vertex is −b ÷ 2a. Substitute that value back into the equation to get the y-coordinate. The vertex is the lowest point when a is positive and the highest point when a is negative.
Yes, when the roots are simple. For x² − 3x + 2 = 0 you can write (x − 1)(x − 2) = 0 and read off x = 1 and x = 2. For messier or non-integer roots, the quadratic formula is more reliable.
If a is zero the equation is no longer quadratic; it becomes the linear equation bx + c = 0, which has a single solution x = −c ÷ b. A quadratic must have a non-zero a for the x² term to exist.