A quadratic equation is any equation that can be written as ax² + bx + c = 0, where a, b and c are numbers and a is not zero. They show up everywhere in algebra, physics and on the SAT, so it is worth being comfortable with all three ways to solve them.
Step zero: get everything on one side
Every method starts with the equation in standard form, with zero on one side. If you have x² = 5x + 6, subtract 5x and 6 from both sides first:
x² − 5x − 6 = 0
Now a = 1, b = −5 and c = −6. Keep the signs with the numbers; that’s where most mistakes start.
Method 1: Factoring
Factoring works when the quadratic splits neatly into two binomials. It is the fastest method when it works.
When a = 1
For x² + bx + c, look for two numbers that multiply to c and add to b.
Example: solve x² + 5x + 6 = 0.
- We need two numbers that multiply to 6 and add to 5. Those are 2 and 3.
- Write the factors: (x + 2)(x + 3) = 0.
- If a product is zero, one of the factors must be zero: x + 2 = 0 or x + 3 = 0.
- So x = −2 or x = −3.
Check: (−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0. ✓
When a is not 1
Example: solve 2x² + 7x + 3 = 0.
- Multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: 6 and 1.
- Split the middle term: 2x² + 6x + x + 3 = 0.
- Group and factor: 2x(x + 3) + 1(x + 3) = 0, which is (2x + 1)(x + 3) = 0.
- So 2x + 1 = 0 or x + 3 = 0, giving x = −1/2 or x = −3.
Don’t divide by x
A common trap: for x² = 5x, it’s tempting to divide both sides by x and get x = 5. But that loses a solution. Instead, write x² − 5x = 0 and factor: x(x − 5) = 0, so x = 0 or x = 5. Dividing by a variable can throw away the answer where that variable is zero.
Method 2: Completing the square
Completing the square works for every quadratic. It is also how the quadratic formula is derived, and it is the method you need to rewrite a parabola in vertex form.
Example: solve x² + 6x − 7 = 0.
- Move the constant to the other side: x² + 6x = 7.
- Take half of b and square it: half of 6 is 3, and 3² = 9.
- Add 9 to both sides: x² + 6x + 9 = 16.
- The left side is now a perfect square: (x + 3)² = 16.
- Take the square root of both sides, remembering ±: x + 3 = ±4.
- Solve both: x = −3 + 4 = 1, or x = −3 − 4 = −7.
Check x = 1: 1 + 6 − 7 = 0. ✓ Check x = −7: 49 − 42 − 7 = 0. ✓
If a is not 1, divide every term by a first, then follow the same steps.
Method 3: The quadratic formula
The quadratic formula solves any quadratic equation:
x = (−b ± √(b² − 4ac)) / 2a
Example: solve 2x² − 4x − 3 = 0.
- Identify the values: a = 2, b = −4, c = −3.
- Calculate the discriminant: b² − 4ac = (−4)² − 4(2)(−3) = 16 + 24 = 40.
- Substitute: x = (4 ± √40) / 4.
- Simplify √40 = 2√10, so x = (4 ± 2√10) / 4 = (2 ± √10) / 2.
- As decimals: x ≈ 2.58 or x ≈ −0.58.
Notice step 3: b is −4, so −b is +4. Getting this sign wrong is the single most common error with the formula.
What the discriminant tells you
The expression under the square root, b² − 4ac, is called the discriminant. You can calculate it first to know what kind of answers to expect:
| Discriminant | Number of real solutions | What the graph does |
|---|---|---|
| Positive | Two | Crosses the x-axis twice |
| Zero | One (a repeated root) | Touches the x-axis once, at the vertex |
| Negative | None | Never touches the x-axis |
Example: x² + 2x + 5 = 0 has discriminant 2² − 4(1)(5) = 4 − 20 = −16. It’s negative, so there are no real solutions. (In Algebra II you’ll learn these have complex solutions.)
The SAT likes questions that ask how many solutions an equation has. The discriminant answers them in one line.
Which method should you use?
- Try factoring first if the numbers are small and a is 1. If you can’t find the pair of numbers in about thirty seconds, move on.
- Use the quadratic formula when the numbers are messy, a isn’t 1, or the problem asks for decimal answers.
- Complete the square when the problem asks for vertex form, the maximum or minimum value, or when b is even and a is 1.
- Graph it to check your answers: the solutions are where the parabola crosses the x-axis. Our free graphing calculator makes this quick.
Mistakes that cost the most points
- Not setting the equation equal to zero before factoring or using the formula.
- Dropping the negative sign on b or c.
- Forgetting the ± when taking a square root, which loses one solution.
- Dividing only part of the numerator by 2a. The whole expression −b ± √(b² − 4ac) is divided by 2a.
- Dividing both sides by x, which loses the solution x = 0.
One to try yourself: solve x² − x − 12 = 0 by factoring, then check your answer with the quadratic formula. (Answer: x = 4 or x = −3.)
Your turn
Solve x² − 9x + 20 = 0 by factoring.
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x = 4 or x = 5. The numbers −4 and −5 multiply to 20 and add to −9.
Solve 3x² + 10x − 8 = 0.
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x = ⅔ or x = −4. ac = −24, and 12 and −2 add to 10: 3x² + 12x − 2x − 8 = 3x(x + 4) − 2(x + 4) = (3x − 2)(x + 4).
Solve x² + 8x + 3 = 0 by completing the square.
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x = −4 ± √13. Move the 3, add 16 to both sides: (x + 4)² = 13.
Solve 2x² + 3x − 4 = 0 with the quadratic formula.
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x = (−3 ± √41)/4, about 0.85 or −2.35. The discriminant is 9 + 32 = 41.
How many real solutions does 4x² − 12x + 9 = 0 have?
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One. The discriminant is 144 − 144 = 0. The equation is (2x − 3)² = 0, so x = 3/2.
A ball’s height in meters is h = −5t² + 20t + 25 after t seconds. When does it hit the ground?
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After 5 seconds. Set h = 0 and divide by −5: t² − 4t − 5 = 0, so (t − 5)(t + 1) = 0. A negative time doesn’t make sense, so t = 5.
For the factoring methods in more depth, see Factoring Polynomials. For how quadratics appear on the SAT, see the SAT Math guide.