04 // Interactive Formula Compendium // CALCULUS-ALGEBRAx = (−b ± √(b² − 4ac)) / (2a)
Quadratic Formula — Roots, Discriminant & Vertex Reference Sheet
The quadratic formula x = (−b ± √(b² − 4ac)) / (2a) determines the exact real or complex roots of the univariate quadratic equation ax² + bx + c = 0 for any non-zero leading coefficient a ≠ 0.
Formula Reference
Main Formula
x = (−b ± √(b² − 4ac)) / (2a)
Rearranged Forms
Δ = b² − 4ac (Discriminant)
x_vertex = −b / (2a), y_vertex = c − b² / (4a)
x₁ + x₂ = −b / a, x₁ · x₂ = c / a (Vieta's Formulas)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| x₁ | Principal Root (x₁ = (−b + √Δ) / 2a) | dimensionless |
| a | Quadratic Coefficient (a ≠ 0) | 1 |
| b | Linear Coefficient | 1 |
| c | Constant Term (y-intercept) | 1 |
Inputs & Parameters
[1]
[1]
[1]
Calculated Result (Principal Root (x₁ = (−b + √Δ) / 2a))
x₁ =3dimensionless
Formula Notes & Background
- Divide ax² + bx + c = 0 through by a ≠ 0 to normalize the leading term: x² + (b/a)x = −c/a.
- Complete the square by adding (b / 2a)² to both sides: (x + b / 2a)² = (b² − 4ac) / (4a²).
- Take the square root of both sides and isolate x to obtain x = (−b ± √(b² − 4ac)) / (2a).
Step-by-Step Examples
#1: Solve x² − 5x + 6 = 0 for its two real intersections with the x-axis.
a = 1b = −5c = 6
x = (−(−5) ± √((−5)² − 4(1)(6))) / (2 × 1) = (5 ± √(25 − 24)) / 2
→ x₁ = 3, x₂ = 2 (Δ = 1 > 0, two distinct real roots)
#2: Determine the roots of 2x² + 4x − 6 = 0.
a = 2b = 4c = −6
x = (−4 ± √(16 − 4(2)(−6))) / 4 = (−4 ± √64) / 4 = (−4 ± 8) / 4
→ x₁ = 1, x₂ = −3 (Δ = 64)
Was this tool helpful?
Frequently Asked Questions
What does the sign of the discriminant Δ = b² − 4ac tell you?+
If Δ > 0, the parabola crosses the x-axis at two distinct real roots; if Δ = 0, it touches the axis at one repeated real root x = −b/(2a); if Δ < 0, it has two complex conjugate roots.
How do you avoid catastrophic cancellation in floating-point code when b² ≫ |4ac|?+
Compute q = −0.5 × (b + sign(b)√Δ), then evaluate x₁ = q / a and x₂ = c / q so two nearly equal numbers are never subtracted.
How does Vieta's formula verify quadratic roots quickly?+
The sum of the roots must equal −b/a and the product of the roots must equal c/a.