Quadratic equations: standard form and how to solve them
A quadratic equation is an equation where the highest power of the unknown is 2. Its standard form is ax² + bx + c = 0, where a, b and c are numbers and a is not 0. A quadratic equation has up to two solutions, called roots, and you can find them by factoring, by taking square roots, by completing the square or with the quadratic formula, x = (−b ± √(b² − 4ac)) / 2a.
What is a quadratic equation?
The word quadratic comes from the Latin for square, because the defining term is x². These are all quadratic equations:
- x² − 5x − 14 = 0, where a = 1, b = −5 and c = −14
- 3x² − 48 = 0, where b = 0
- 2x² + 7x = 0, where c = 0
- 7x² = 9 + x, which becomes 7x² − x − 9 = 0 once everything is on one side
These are not: 4x + 1 = 0 has no x² term, so it is linear, and x³ − x = 0 has an x³ term, so it is a cubic. The equation must contain x², and nothing higher.
Standard form, vertex form and factored form
The same quadratic can be written three ways. Each one shows you something different about its graph, a U-shaped curve called a parabola.
| Form | Looks like | What it shows you |
|---|---|---|
| Standard form | y = ax² + bx + c | The y-intercept, c, and the values to use in the quadratic formula |
| Vertex form | y = a(x − h)² + k | The vertex (turning point) at (h, k) |
| Factored form | y = a(x − r)(x − s) | The roots, x = r and x = s, where the graph crosses the x-axis |
For example, y = x² − 5x − 14 is in standard form. In factored form it is y = (x − 7)(x + 2), so it crosses the x-axis at 7 and −2. In vertex form it is y = (x − 2.5)² − 20.25, so its lowest point is (2.5, −20.25).
Here x is the input and y is the value of x² − 5x − 14. Solving the quadratic equation x² − 5x − 14 = 0 means finding where y = 0, which is where the curve crosses the x-axis. The Quadratic Graph Calculator draws any quadratic and marks its roots and vertex.
How to solve a quadratic equation: 4 ways
First, rearrange so one side is 0. Then choose a method. Each one works on some equations better than others.
| Method | Best when | Example |
|---|---|---|
| Factoring | The equation factors with whole numbers | x² − 5x − 14 = 0 |
| Square roots | There is no x term (b = 0), or it is already a perfect square | 3x² − 48 = 0 |
| Completing the square | a = 1 and b is even, or you need the vertex too | x² + 6x − 7 = 0 |
| Quadratic formula | Always works, and is the only easy choice when it will not factor | 3x² − 4x − 2 = 0 |
1. Solving by factoring
Write the quadratic as two brackets multiplied together. If two things multiply to give 0, one of them must be 0, so each bracket gives one solution.
Solve x² − 5x − 14 = 0. Find two numbers that multiply to −14 and add to −5. They are −7 and 2.
So x − 7 = 0 or x + 2 = 0, which gives x = 7 or x = −2.
Factoring is fastest when it works, but most quadratics do not factor with whole numbers. If you cannot find the pair of numbers in about a minute, switch to the formula. Try factoring one now, then press Mark to check it in the course.
Factor a quadratic into two brackets.
Press Mark to have your answer marked in the Algebra course.
2. Solving by taking square roots
If there is no x term, get x² on its own and square root both sides. Remember that a positive number has two square roots.
Solve 3x² − 48 = 0.
So x = 4 or x = −4, often written x = ±4. The same idea works for a squared bracket: (x − 3)² = 25 gives x − 3 = ±5, so x = 8 or x = −2.
3. Solving by completing the square
Rewrite the quadratic as a squared bracket plus a number, then use square roots. For x² + bx, the bracket is (x + b/2)², and you subtract (b/2)² to keep the value the same.
Solve x² + 6x − 7 = 0. Half of 6 is 3, so:
So x + 3 = ±4, which gives x = 1 or x = −7. Our guide to completing the square covers harder cases, such as when a is not 1.
4. Solving with the quadratic formula
The quadratic formula solves every quadratic equation, including ones that will not factor.
How to use the quadratic formula, step by step
This is the walk-through most people are looking for when they search for how to use the quadratic equation. Take 3x² − 4x − 2 = 0.
- Make one side 0. It already is.
- Read off a, b and c, with their signs. a = 3, b = −4, c = −2.
- Work out the discriminant, b² − 4ac: (−4)² − 4 × 3 × (−2) = 16 + 24 = 40.
- Substitute into the formula.
- Work out both answers, one with + and one with −.
So x = 1.72 or x = −0.387 to 3 significant figures. In exact form, √40 = 2√10, so x = (2 ± √10)/3.
Try one yourself. Use the formula to find both solutions, then press Mark to check them in the course.
Use the quadratic formula to find both roots.
Press Mark to have your answer marked in the Algebra course.
For more practice with the formula alone, including what the vertex tells you, see the quadratic formula calculator guide.
The discriminant of a quadratic equation
The part under the square root, b² − 4ac, is called the discriminant. You can work it out first to see how many real solutions to expect.
| Discriminant | Number of real solutions | Example | Graph |
|---|---|---|---|
| Positive | Two different solutions | x² − 10x + 21 = 0: 100 − 84 = 16, so x = 3 or 7 | Crosses the x-axis twice |
| Zero | One repeated solution | x² − 6x + 9 = 0: 36 − 36 = 0, so x = 3 | Touches the x-axis once |
| Negative | No real solutions | x² + 4x + 7 = 0: 16 − 28 = −12 | Never reaches the x-axis |
When the discriminant is a perfect square, such as 16 or 81, the equation factors with whole numbers. That is a quick way to decide whether factoring is worth trying. A negative discriminant has no real square root, so the solutions are complex numbers, which you meet in Algebra 2 in the US or A-level Further Maths in the UK.
Quadratic equation examples
Example 1: rearrange first
Which equation shows the quadratic formula used correctly to solve 7x² = 9 + x for x?
Subtract 9 and x from both sides to get 7x² − x − 9 = 0. So a = 7, b = −1 and c = −9. The correct substitution is:
That gives x = 1.21 or x = −1.06 to 3 significant figures. The usual wrong answers use b = 1 instead of −1, or c = 9 instead of −9, because the equation was not rearranged first.
Example 2: an area problem
A rectangular garden is 3 m longer than it is wide, and its area is 70 m². Find its width and length.
Let the width be w, so the length is w + 3.
w = −10 or w = 7. A width cannot be negative, so the garden is 7 m wide and 10 m long. Check: 7 × 10 = 70.
Example 3: when does a ball land?
A ball is thrown upward from a height of 4 feet. Its height after t seconds is h = −16t² + 40t + 4. When does it hit the ground?
It hits the ground when h = 0. This does not factor, so use the formula with a = −16, b = 40 and c = 4. The discriminant is 40² − 4(−16)(4) = 1600 + 256 = 1856.
That gives t = −0.096 or t = 2.60. The negative time is before the throw, so the ball lands after 2.60 seconds. US textbooks use feet and 16, half of gravity's 32 ft/s². In metres, the same kind of equation uses 4.9 instead.
Common mistakes
- Not making one side 0 first. x² + 3x = 10 is not ready for factoring or the formula until it is x² + 3x − 10 = 0.
- Losing the sign of b or c. In 3x² − 4x − 2 = 0, b is −4, so −b is +4.
- Squaring a negative b wrongly. (−4)² is +16. Type it on a calculator with brackets.
- Dividing only part of the top by 2a. The whole of −b ± √(b² − 4ac) goes over 2a.
- Forgetting the negative square root. x² = 16 means x = 4 or x = −4.
- Dividing both sides by x. In 2x² + 7x = 0, dividing by x loses the solution x = 0. Factor instead: x(2x + 7) = 0, so x = 0 or x = −3.5.
Is the quadratic formula on the formula sheet?
It depends on the test. This was checked in November 2026.
| Test | Quadratic formula given? |
|---|---|
| SAT (US) | No. The reference sheet in College Board's official practice tests lists only geometry formulas and facts |
| GCSE Maths Higher (UK: AQA, Edexcel, OCR) | Yes, on the Higher formulae sheet |
| GCSE Maths Foundation (UK) | Not on the Foundation sheet. Foundation papers solve quadratics by factorising |
So for the SAT, learn the formula by heart. For GCSE Higher, it is printed for you, but you still need to rearrange the equation and substitute carefully.
Quadratic equation questions with answers
1. Solve x² + x − 12 = 0 by factoring.
(x + 4)(x − 3) = 0, so x = −4 or x = 3
2. Solve 5x² = 45.
3. Solve x² − 4x − 1 = 0, giving exact answers.
x − 2 = ±√5, so x = 2 + √5 or x = 2 − √5
4. Solve 2x² + 5x − 4 = 0, giving answers to 2 decimal places.
x = (−5 ± √57) ÷ 4, so x = 0.64 or x = −3.14
5. How many real solutions does 4x² − 12x + 9 = 0 have?
One repeated solution, x = 1.5, because 4x² − 12x + 9 = (2x − 3)².
6. The product of two consecutive positive integers is 132. Find them.
n is positive, so the integers are 11 and 12.
Practise quadratics in a short timed test
Quadratics turn up all through Algebra 1 and Algebra 2. Pre-Algebra and Algebra 1 and Geometry, Algebra 2 and SAT Math each give you a short timed practice test every day, with 8 questions worth 25 to 35 points. In the UK, the same practice opens as GCSE Maths Higher. To learn the topic from the start, the Algebra course builds from expanding brackets to factoring and the quadratic formula.
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