To solve a quadratic equation, first rearrange it into standard form, ax² + bx + c = 0, then use one of four methods: factoring, taking square roots, completing the square, or the quadratic formula. The quadratic formula, x = (−b ± √(b² − 4ac)) ÷ 2a, works for every quadratic, while factoring is faster when the equation factors neatly. Most quadratics have two solutions, and you can check each by substituting it back into the original equation.
Step zero: get the equation into standard form
A quadratic equation has x² as its highest power. Before choosing a method, move every term to one side so the other side is zero:
ax² + bx + c = 0, where a ≠ 0
For example, 2x² = 3x + 5 becomes 2x² − 3x − 5 = 0, so a = 2, b = −3 and c = −5. Keep the signs with the coefficients; most errors in the quadratic formula come from dropping a minus sign here.
Method 1: Factoring
Factoring rewrites the quadratic as a product of two brackets. If a product equals zero, one of the factors must be zero (the zero product property).
Example: when a = 1
Solve x² − 5x + 6 = 0.
- Find two numbers that multiply to c = 6 and add to b = −5. They are −2 and −3.
- Factor: (x − 2)(x − 3) = 0
- Set each factor to zero: x − 2 = 0 or x − 3 = 0
- Solutions: x = 2 or x = 3
Check: 2² − 5(2) + 6 = 4 − 10 + 6 = 0 ✓ and 3² − 5(3) + 6 = 9 − 15 + 6 = 0 ✓
Example: when a ≠ 1
Solve 2x² + 7x + 3 = 0.
- Multiply a × c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: they are 6 and 1.
- Split the middle term: 2x² + 6x + x + 3 = 0
- Group: 2x(x + 3) + 1(x + 3) = 0
- Factor: (2x + 1)(x + 3) = 0
- Solutions: x = −1/2 or x = −3
Check x = −1/2: 2(1/4) + 7(−1/2) + 3 = 0.5 − 3.5 + 3 = 0 ✓
Example: no constant term
Solve x² − 4x = 0. Factor out x: x(x − 4) = 0, so x = 0 or x = 4.
Don't divide both sides by x here. That throws away the solution x = 0.
Method 2: Taking square roots
When there's no x term (b = 0), isolate x² and take the square root of both sides. Remember the ±.
Solve 3x² − 48 = 0.
- Add 48: 3x² = 48
- Divide by 3: x² = 16
- Square root: x = ±4, so x = 4 or x = −4
The same method works on equations like (x − 1)² = 9: x − 1 = ±3, so x = 4 or x = −2.
Method 3: Completing the square
Completing the square turns the left side into a perfect square, so you can use square roots. It's also how the quadratic formula is derived.
Solve x² − 4x − 1 = 0.
- Move the constant: x² − 4x = 1
- Take half of b and square it: half of −4 is −2, and (−2)² = 4
- Add 4 to both sides: x² − 4x + 4 = 5
- Write the left side as a square: (x − 2)² = 5
- Square root: x − 2 = ±√5
- Solutions: x = 2 + √5 or x = 2 − √5 (about 4.236 and −0.236)
If a isn't 1, divide every term by a first so the x² term has a coefficient of 1.
Method 4: The quadratic formula
The quadratic formula solves any quadratic in standard form:
x = (−b ± √(b² − 4ac)) ÷ 2a
The whole numerator, −b and the square root together, is divided by 2a.
Example with whole-number answers
Solve 2x² − 3x − 5 = 0, where a = 2, b = −3, c = −5.
- Discriminant: b² − 4ac = (−3)² − 4(2)(−5) = 9 + 40 = 49
- Square root: √49 = 7
- Formula: x = (3 ± 7) ÷ 4
- Solutions: x = 10 ÷ 4 = 5/2 or x = −4 ÷ 4 = −1
Check x = 5/2: 2(25/4) − 3(5/2) − 5 = 12.5 − 7.5 − 5 = 0 ✓
Example with irrational answers
Solve 3x² + 2x − 2 = 0, where a = 3, b = 2, c = −2.
- Discriminant: 2² − 4(3)(−2) = 4 + 24 = 28
- Simplify the root: √28 = √(4 × 7) = 2√7
- Formula: x = (−2 ± 2√7) ÷ 6
- Simplify by dividing every term by 2: x = (−1 ± √7) ÷ 3
As decimals, that's about 0.549 and −1.215.
The discriminant: how many solutions?
The expression under the square root, b² − 4ac, is called the discriminant. It tells you how many real solutions to expect before you finish solving.
| Discriminant | Real solutions | Graph of y = ax² + bx + c | Example |
|---|---|---|---|
| Positive | Two different | Crosses the x-axis twice | 2x² − 3x − 5 = 0 (discriminant 49) |
| Zero | One repeated | Touches the x-axis once | x² − 6x + 9 = 0 (discriminant 0) |
| Negative | None (two complex) | Never touches the x-axis | x² + 2x + 5 = 0 (discriminant −16) |
For x² − 6x + 9 = 0: the discriminant is 36 − 36 = 0, and x = 6 ÷ 2 = 3. The equation is (x − 3)² = 0.
For x² + 2x + 5 = 0: the discriminant is 4 − 20 = −16. There are no real solutions. Using the imaginary unit i (where i² = −1), the complex solutions are x = (−2 ± 4i) ÷ 2 = −1 ± 2i.
A positive discriminant that's a perfect square (like 49) means the equation factors with whole numbers or fractions. A positive discriminant that isn't (like 28) means the solutions are irrational.
Which method should you use?
| Situation | Best method |
|---|---|
| Small numbers that obviously factor | Factoring |
| No x term (b = 0) | Square roots |
| No constant term (c = 0) | Factor out x |
| a = 1 and b is even | Completing the square works neatly |
| Anything else, or you're not sure | Quadratic formula |
When in doubt, compute the discriminant first. If it's a perfect square, try factoring; otherwise, go straight to the formula.
A word problem
A rectangle's length is 3 cm more than its width, and its area is 40 cm². Find the dimensions.
- Let the width be w. Then the length is w + 3.
- Area: w(w + 3) = 40, so w² + 3w − 40 = 0
- Factor: two numbers that multiply to −40 and add to 3 are 8 and −5, so (w + 8)(w − 5) = 0
- Solutions: w = −8 or w = 5
- A width can't be negative, so w = 5 cm and the length is 8 cm. Check: 5 × 8 = 40 ✓
Word problems often produce one solution that doesn't make sense in context. Say why you reject it.
How to check your answers
- Substitute. Put each solution back into the original equation. Both sides should match.
- Use sum and product. For ax² + bx + c = 0, the two solutions add to −b/a and multiply to c/a. For 2x² − 3x − 5 = 0: 5/2 + (−1) = 3/2 = −(−3)/2 ✓ and (5/2)(−1) = −5/2 = c/a ✓.
- Sketch the graph. The real solutions are where the parabola crosses the x-axis.
Common mistakes
- Forgetting to set the equation equal to zero before factoring or using the formula
- Getting the sign of −b wrong when b is negative (if b = −3, then −b = 3)
- Dividing only the square root by 2a instead of the whole numerator
- Losing the ± and finding only one solution
- Dividing both sides by x and losing x = 0
Practice and check your work
Work each problem by hand first, then compare. An equation solver that shows exact roots, steps and a substitution check is useful for spotting where your work went off track. Use AI step-by-step explanations as a study aid, check them against your own work, and follow your class rules on calculators and AI; our guide on using AI without violating academic integrity covers where the lines usually are.
Solve for x CalculatorExact solutions to equations, step by stepQuadratics come up again in calculus. The vertex of a parabola is where its slope is zero, which you can find with the power rule in our guide to derivative rules.
Key takeaways
- Rewrite the equation as ax² + bx + c = 0 before choosing a method.
- Factor when it's quick, use square roots when b = 0, and use the quadratic formula for everything else.
- The discriminant b² − 4ac tells you whether there are two, one or no real solutions.
- Check every answer by substitution, and reject solutions that don't fit a word problem's context.
