Answer: Enter the coefficients a, b and c to get both roots of a quadratic equation, including complex solutions and the discriminant.
Solve quadratic equations instantly.
This quadratic equation solver finds the roots of any equation ax² + bx + c = 0 — real or complex — showing the discriminant, vertex, and axis of symmetry alongside the answers. All computation is local and instant.
Quadratics are the first functions with genuinely interesting behavior: they curve, they cross or miss the x-axis, and their solutions connect algebra to geometry through one number — the discriminant.
For ax² + bx + c = 0 with a ≠ 0, the roots are x = (−b ± √(b² − 4ac)) ÷ 2a. The expression under the root, D = b² − 4ac, is the discriminant, and it predicts the solution type before you compute anything.
If D > 0 there are two distinct real roots; if D = 0 there is exactly one repeated real root; if D < 0 the two roots are complex conjugates of the form p ± qi. This works because the parabola either crosses the x-axis twice, touches it once at its vertex, or misses it entirely.
The table walks three equations through the discriminant classification computed exactly. For x² − 3x + 2: D = 9 − 8 = 1, roots (3 ± 1) ÷ 2 = 2 and 1. For x² + 4x + 4: D = 16 − 16 = 0, one root −2. For x² + 2x + 5: D = 4 − 20 = −16, roots −1 ± 2i (since √−16 = 4i).
Verify any root by substitution: putting x = 2 into x² − 3x + 2 gives 4 − 6 + 2 = 0. A solution that doesn't substitute back to zero isn't a solution.
| Equation | a, b, c | Discriminant | Roots |
|---|---|---|---|
| x² − 3x + 2 = 0 | 1, −3, 2 | 1 (D > 0) | x = 2, x = 1 |
| x² + 4x + 4 = 0 | 1, 4, 4 | 0 (D = 0) | x = −2 (repeated) |
| x² + 2x + 5 = 0 | 1, 2, 5 | −16 (D < 0) | x = −1 ± 2i |
Every quadratic graphs as a parabola. Its vertex sits at x = −b ÷ 2a, with y-value c − b² ÷ 4a (equivalently, substitute −b/2a into the equation). The vertical line through the vertex is the axis of symmetry, and the parabola opens upward if a > 0, downward if a < 0.
This makes quadratics the workhorse of optimization: the vertex is the maximum or minimum of the quantity modeled. Projectile height h(t) = −16t² + v₀t + h₀ (with t in seconds and h in feet, using 32 ft/s² gravitational acceleration halved) peaks at t = −v₀ ÷ (2 × −16) = v₀ ÷ 32 — the vertex formula doing physics.
Three solution methods, one answer. Factoring finds numbers multiplying to ac and summing to b — fast when integers exist, impossible when they don't. Completing the square rewrites ax² + bx + c as a(x + b/2a)² + (c − b²/4a), revealing the vertex form directly. The quadratic formula is completing the square done once, symbolically, for all cases.
Historically this generality mattered: the formula handles irrational and complex roots that factoring misses, and its discriminant test classifies the answer in advance. That's why the formula is the default and the other two methods are shortcuts for special cases.
['Quadratics model anything with constant acceleration or area scaling. Projectile motion: h(t) = −16t² + v₀t + h₀ (feet, seconds) describes a ball thrown at 32 ft/s from 5 feet up — peak at t = 32 ÷ 32 = 1 second, landing when −16t² + 32t + 5 = 0, i.e. t = (32 + √(1024+320)) ÷ 32 ≈ 2.14 seconds. Vertex and roots, one formula each.', 'Business and geometry: revenue often peaks mid-range (price too high or too low both lose) — a quadratic whose vertex is the optimal price. And areas are products of dimensions: fencing 40 feet of pen against a barn wall with width w gives area w(40 − 2w), maximized at the vertex w = 10, area 200 square feet.', 'The complex roots are not wasted either: they describe oscillations (damped systems, AC circuits) where a negative discriminant means no real equilibrium crossing but a decaying sinusoid instead. The math you meet in algebra class is the same math that tunes suspension dampers and audio filters.']
What is the quadratic formula?
For ax² + bx + c = 0, x = (−b ± √(b² − 4ac)) ÷ 2a. It always gives the roots — real or complex — for any quadratic with a ≠ 0.
What does the discriminant tell you?
D = b² − 4ac predicts the root type: D > 0 means two distinct real roots, D = 0 means one repeated real root, and D < 0 means two complex conjugate roots. Geometrically it says whether the parabola crosses, touches, or misses the x-axis.
When does a quadratic have no real solutions?
When the discriminant is negative (b² − 4ac < 0). The parabola never reaches the x-axis. Example: x² + 2x + 5 has D = 4 − 20 = −16, with roots −1 ± 2i.
How do I find the vertex of a parabola?
The vertex is at x = −b ÷ 2a; substitute to get y (or use y = c − b² ÷ 4a). For x² − 3x + 2, the vertex is at x = 1.5, y = −0.25 — the minimum point.
Can I solve quadratics by factoring instead?
Yes, when rational roots exist: x² − 3x + 2 = (x − 2)(x − 1). Factoring is faster but only works for special cases; the quadratic formula handles everything, including irrational and complex roots.