Standard form first
A quadratic equation is any equation you can write as:
Before doing anything else, rearrange so one side is zero. Every method below assumes it. If you're staring at x² = 5x − 6, rewrite it as x² − 5x + 6 = 0 first — skipping this step is the most common reason an otherwise correct method gives the wrong answer.
Method 1: factoring
Fastest when it works. You're looking for two numbers that multiply to c and add to b (when a = 1).
- Find the pair. Which two numbers multiply to 6 and add to 5? That's 2 and 3.
- Write the factors. (x + 2)(x + 3) = 0
- Apply the zero-product rule. If two things multiply to zero, at least one must be zero. So x + 2 = 0 or x + 3 = 0.
- Solve each. x = −2 or x = −3
Check by substituting: (−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0 ✓
For something like 2x² + 3x − 2, look for two numbers that multiply to a×c = −4 and add to b = 3. That's 4 and −1, so split the middle term: 2x² + 4x − x − 2, then group: 2x(x + 2) − 1(x + 2) = (2x − 1)(x + 2).
Factoring's weakness: it only works cleanly when the roots are rational. Give it five to ten seconds, and if no obvious pair appears, switch methods rather than grinding.
Method 2: the quadratic formula
Never fails. Works for every quadratic, rational roots or not.
Take 2x² + 3x − 2 = 0, so a = 2, b = 3, c = −2.
- Compute the discriminant. b² − 4ac = 9 − 4(2)(−2) = 9 + 16 = 25. Watch the double negative here — it's where most errors happen.
- Take the root. √25 = 5
- Build both solutions. x = (−3 ± 5) / 4
- Split them out. x = 2/4 = 1/2, and x = −8/4 = −2
Check the first: 2(1/2)² + 3(1/2) − 2 = 0.5 + 1.5 − 2 = 0 ✓
Write down a, b and c explicitly before substituting — most formula errors are sign errors on b or c, not arithmetic. And keep the whole −b together: if b = −7, then −b = +7.
Method 3: completing the square
Slower for pure solving, but it's the method that matters most later — it's how you find the vertex of a parabola, derive the quadratic formula itself, and handle conic sections and certain integrals.
- Move the constant. x² + 6x = −5
- Halve the x-coefficient, then square it. Half of 6 is 3; 3² = 9. Add 9 to both sides: x² + 6x + 9 = 4
- Factor the perfect square. (x + 3)² = 4
- Take the square root of both sides. x + 3 = ±2 — the ± is essential, and forgetting it loses one of your two roots.
- Solve. x = −1 or x = −5
Check: (−1)² + 6(−1) + 5 = 1 − 6 + 5 = 0 ✓ and (−5)² + 6(−5) + 5 = 25 − 30 + 5 = 0 ✓
As a bonus, the completed form (x + 3)² − 4 = 0 tells you the parabola's vertex sits at (−3, −4) — information the other methods don't hand you.
Method 4: graphing
Plot y = ax² + bx + c and read off where the curve crosses the x-axis. Those crossings are the solutions, also called roots or zeros.
This is the least precise method — you'll rarely read an exact irrational root off a graph — but it's the most illuminating, because it shows you why a quadratic has two solutions, one, or none:
- Curve crosses the axis twice → two real solutions
- Curve just touches the axis → one repeated solution
- Curve never reaches the axis → no real solutions
Use graphing to check an answer you got algebraically, or when a question asks you to interpret a situation rather than produce exact values.
The discriminant: how many solutions?
The part under the root sign, b² − 4ac, decides everything before you do any work:
- b² − 4ac > 0 — two distinct real solutions
- b² − 4ac = 0 — exactly one repeated real solution
- b² − 4ac < 0 — no real solutions (two complex ones)
Worth computing first whenever a question asks "how many solutions" or "for what values of k does this have real roots" — you can answer without solving at all. And a perfect-square discriminant (like our 25) tells you the equation would have factored, so next time you'd save time by factoring instead.
Which method should you use?
- Is c = 0? Factor out x immediately. x² − 5x = 0 becomes x(x − 5) = 0, giving x = 0 and x = 5. No formula needed.
- Is b = 0? Isolate and square-root. 2x² − 18 = 0 → x² = 9 → x = ±3.
- Do two obvious numbers work? Factor. Give yourself about ten seconds to spot them.
- Anything else? Quadratic formula. It always works, and at that point it's the fastest reliable route.
- Asked for a vertex, a maximum, or a minimum? Complete the square, regardless of the above.
Whichever route you take, substitute at least one root back into the original equation. It costs fifteen seconds and catches nearly every sign error — see our guide to checking your answers for more of these.
Practice until it's automatic
Scan a quadratic with Aheadia for the full working — or turn on Study Mode and it'll walk you through a similar one first, then hand yours back for you to try.
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