The Quadratic Formula
The method that always works, and what the discriminant predicts.
The explanation
The quadratic formula solves every quadratic, factorable or not.
For ax² + bx + c = 0:
x = (−b ± √(b² − 4ac)) / (2a)
Write the equation in standard form first and identify a, b, c *with their signs*. In 2x² − 5x − 3 = 0, b is −5, not 5.
The part under the root, b² − 4ac, is the discriminant, and it predicts the answer before you finish:
- positive → two real solutions
- zero → one repeated solution
- negative → no real solutions
Substitute carefully with parentheses. Most errors here are sign errors, not formula errors.
Derived by completing the square on ax² + bx + c = 0, the formula gives both roots for any a ≠ 0 over ℝ or ℂ.
The discriminant Δ = b² − 4ac determines the nature of the roots: Δ > 0 gives two distinct real roots (and if Δ is a perfect square with rational coefficients, they are rational and the quadratic factors over ℚ); Δ = 0 gives one real root of multiplicity 2, where the parabola is tangent to the x-axis; Δ < 0 gives a conjugate pair of complex roots p ± qi.
Vieta's formulas follow directly from the two roots: their sum is −b/a and their product is c/a. These give a quick check on any computed pair and are the fastest way to reconstruct a quadratic from its roots.
Numerically, the formula suffers catastrophic cancellation when b² ≫ 4ac and the signs align, which is why computational implementations use the algebraically equivalent form 2c/(−b ∓ √Δ) for one of the roots.
Worked example
Solve 3x² + 2x − 4 = 0.
- a = 3, b = 2, c = −4.
- Discriminant: 2² − 4(3)(−4) = 4 + 48 = 52.
- x = (−2 ± √52)/6 and √52 = 2√13.
- x = (−2 ± 2√13)/6 = (−1 ± √13)/3.
Answer: x = (−1 + √13)/3 or x = (−1 − √13)/3
Common mistakes
- Dropping the sign of b or c when identifying coefficients.
- Dividing only the radical by 2a instead of the whole numerator.