Quadratic Formula Calculator
Solve ax² + bx + c = 0 for both roots with discriminant and vertex.
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The Quadratic Formula
The quadratic formula solves any quadratic equation of the form ax² + bx + c = 0: x = (−b ± √(b²−4ac)) / (2a). It gives up to two solutions (roots) for x. The formula works for all quadratics, even when factoring is difficult or impossible.
The Discriminant
The discriminant Δ = b² − 4ac determines the nature of the roots: if Δ > 0, there are two distinct real roots; if Δ = 0, there is exactly one real root (a repeated root); if Δ < 0, there are no real roots (two complex conjugate roots).
Finding the Vertex
The parabola y = ax² + bx + c has its vertex at x = −b/(2a). The y-coordinate of the vertex is found by substituting this x back into the equation. The axis of symmetry is the vertical line x = −b/(2a).
Factoring vs. Quadratic Formula
Factoring is faster when the roots are integers, but the quadratic formula always works. Example: x² − 5x + 6 = 0 factors as (x−2)(x−3) = 0, giving x = 2 or x = 3. The formula gives the same: x = (5 ± √(25−24))/2 = (5 ± 1)/2 = 3 or 2.
Applications
Quadratic equations model projectile motion (height = −½gt² + v₀t + h₀), optimization problems (maximizing area), engineering design (parabolic arches and reflectors), and economics (revenue = price × demand).
Worked Example: Solving 2x² + 3x − 2 = 0
Identify coefficients: a = 2, b = 3, c = −2.
Step 1 — Calculate the discriminant: Δ = b² − 4ac = 9 − 4(2)(−2) = 9 + 16 = 25. Since Δ > 0, there are two distinct real roots.
Step 2 — Apply the formula: x = (−3 ± √25) / (2×2) = (−3 ± 5) / 4.
Step 3 — Solve both roots: x₁ = (−3 + 5) / 4 = 2/4 = 0.5; x₂ = (−3 − 5) / 4 = −8/4 = −2.
Verification: 2(0.5)² + 3(0.5) − 2 = 0.5 + 1.5 − 2 = 0 ✓. And 2(−2)² + 3(−2) − 2 = 8 − 6 − 2 = 0 ✓.
Frequently Asked Questions
x = (−b ± √(b²−4ac)) / (2a). It solves any equation of the form ax² + bx + c = 0.
b² − 4ac. If > 0: two real roots. If = 0: one real root. If < 0: no real roots (complex roots).
a=1, b=-5, c=6. Discriminant = 25-24=1. x = (5 ± 1)/2 = 3 or 2.
No real number solutions exist. The parabola doesn't cross the x-axis. Solutions are complex numbers involving i (imaginary unit).
The highest or lowest point. x-coordinate = −b/(2a). Substitute back to find y. For x²−5x+6: vertex x = 5/2 = 2.5, y = 6.25−12.5+6 = −0.25.
Use factoring when the roots appear to be small integers — it is faster. Use the quadratic formula when factoring is not obvious or when the discriminant is not a perfect square (giving irrational roots). The formula always works regardless of the values of a, b, and c.
Completing the square is a technique where you rewrite ax² + bx + c in the form a(x + h)² + k. The quadratic formula is derived by completing the square on the general form ax² + bx + c = 0. Both methods are equivalent and yield the same roots.