A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a is not zero. Unlike linear equations, quadratics can have zero, one, or two real solutions — and the quadratic formula handles all three cases at once.
Step 1: Get the equation into standard form
Before doing anything else, move every term to one side so the equation equals zero: ax² + bx + c = 0. If your equation is written as, say, x² - 5x = -6, add 6 to both sides first to get x² - 5x + 6 = 0.
Step 2: Identify a, b, and c
Once in standard form, read off the three coefficients directly. For x² - 5x + 6 = 0: a = 1, b = -5, c = 6. Getting the signs right here is the single most common source of error in the rest of the process, so it's worth writing them out explicitly rather than substituting from memory.
Step 3: Apply the quadratic formula
The quadratic formula is:
x = (−b ± √(b² − 4ac)) / (2a)
This single formula always gives you the solutions — you never need to guess or factor by trial and error, though factoring can sometimes be faster for simple cases.
Step 4: Compute the discriminant first
The expression under the square root, b² − 4ac, is called the discriminant, and it tells you what kind of solutions to expect before you finish the calculation:
- Positive discriminant: two distinct real solutions.
- Zero discriminant: exactly one repeated real solution.
- Negative discriminant: no real solutions (the two solutions are complex numbers).
Worked example 1: Two real solutions
Solve x² - 5x + 6 = 0.
Here a = 1, b = -5, c = 6. The discriminant is (-5)² - 4(1)(6) = 25 - 24 = 1, which is positive, so we expect two real solutions.
Substituting into the formula: x = (5 ± √1) / 2 = (5 ± 1) / 2. That gives x = 3 or x = 2.
Worked example 2: One repeated solution
Solve x² - 4x + 4 = 0.
Here a = 1, b = -4, c = 4. The discriminant is 16 - 16 = 0, so there's exactly one solution: x = -(-4)/2(1) = 2.
Worked example 3: No real solutions
Solve x² + x + 1 = 0.
Here a = 1, b = 1, c = 1. The discriminant is 1 - 4 = -3, which is negative — this equation has no real solutions. Graphically, this means the parabola never crosses the x-axis.
A faster path when factoring is obvious
If you can spot two numbers that multiply to c and add to b (for a = 1 equations), factoring directly can be quicker than the full formula. For x² - 5x + 6 = 0, the numbers -2 and -3 multiply to 6 and add to -5, giving (x - 2)(x - 3) = 0 — the same two solutions found above, reached a different way. When factoring isn't obvious, the quadratic formula always works as a fallback.
Common mistakes specific to quadratics
Two errors show up disproportionately often when solving quadratics. First, forgetting that −b requires flipping the sign of b — if b is already negative, −b becomes positive, and it's easy to lose that flip under time pressure. Second, forgetting that the entire numerator, including the ±√(...) part, gets divided by 2a — not just the −b part. Writing the formula out in full each time, rather than trying to shortcut it from memory, avoids both.
When a equals something other than 1
The examples above all used a = 1 for clarity, but the formula works identically when it doesn't. Solve 2x² + 7x + 3 = 0: here a = 2, b = 7, c = 3. The discriminant is 49 - 24 = 25, a perfect square, so expect clean real solutions. Substituting: x = (-7 ± 5) / 4, giving x = -0.5 or x = -3. The only difference from the a = 1 case is that 2a in the denominator is now 4 instead of 2 — everything else follows the same process.
The discriminant is worth calculating on its own, even before you finish solving — it tells you what shape of answer to expect and catches arithmetic errors early.
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