What makes an equation quadratic
A quadratic equation has the form ax² + bx + c = 0, where a isn't zero. The defining feature is the squared term — it's what makes the graph a parabola instead of a straight line, and it's why a quadratic can have up to two solutions instead of exactly one. Three standard methods solve these equations: factoring, the quadratic formula, and completing the square. They always agree on the answer when applied correctly; they differ in how much work each one takes for a given equation.
Factoring: fastest when it works
Factoring rewrites ax² + bx + c as a product of two binomials, then uses the fact that if a product equals zero, at least one factor must be zero. For x² + 5x + 6 = 0, this factors to (x + 2)(x + 3) = 0, giving solutions x = -2 and x = -3.
Factoring is the fastest method when the equation factors cleanly with integer or simple fractional roots — which is common in textbook problems but not guaranteed in general. The catch: not every quadratic factors nicely over the rational numbers, and spending several minutes hunting for factors that don't exist wastes time better spent on a method guaranteed to work.
The quadratic formula: always works
For any quadratic in the form ax² + bx + c = 0, the solutions are:
x = [-b ± √(b² - 4ac)] / 2a
This formula is derived from completing the square on the general equation, which is why it always produces the correct answer regardless of whether the equation factors nicely. The trade-off is that it involves more arithmetic than factoring when factoring is available, and it's easy to make a sign error plugging in negative values of b or c — writing out each substitution step explicitly, rather than doing it in one line, catches most of those errors.
The expression under the square root, b² - 4ac, is called the discriminant, and it reveals the nature of the solutions before finishing the calculation: positive means two distinct real solutions, zero means exactly one repeated real solution, and negative means the solutions are complex (not real) numbers.
Completing the square: builds understanding, not just an answer
Completing the square rewrites the quadratic as a perfect square trinomial plus a constant, which directly reveals the vertex of the parabola — useful when the goal is graphing or understanding the equation's structure, not just finding the roots. For x² + 6x + 5 = 0: take half of the coefficient on x (which is 3), square it (9), and add and subtract it to keep the equation balanced: x² + 6x + 9 - 9 + 5 = 0, which becomes (x + 3)² - 4 = 0, giving (x + 3)² = 4, so x + 3 = ±2, and x = -1 or x = -5.
This method takes more steps than the quadratic formula for a plain "solve for x" problem, but it's the method underlying the quadratic formula's derivation, and it's the standard technique for converting a quadratic from standard form to vertex form — genuinely useful beyond just solving for roots.
A quick way to choose
- Try factoring first, briefly — if the equation is simple and the factors are obvious within a few seconds, use it.
- If factoring isn't quick, use the quadratic formula. It always works and doesn't require guessing.
- Use completing the square specifically when the problem asks for vertex form, or when the derivation itself is what's being taught, rather than as a default solving method.
A shared root, three different amounts of work
All three methods solve the same equation x² - 4x - 5 = 0 and land on x = 5 and x = -1 — factoring gets there in one line (x - 5)(x + 1) = 0, the quadratic formula takes a few lines of substitution, and completing the square takes the most steps but reveals the vertex (2, -9) along the way. None of the three is "more correct" than the others; picking the right one for the situation is what makes solving quadratics efficient rather than mechanical.