Year 10 · Mathematics · Algebra
Solving quadratic equations
Solve quadratic equations using the null factor law and the quadratic formula, and use the discriminant to decide how many real solutions there are.
Awaiting educator review. This material was drafted by our content team with AI assistance and is waiting for review by a qualified educator. Please check it suits your child before using it. Created 10 October 2026.
Learning goals
Learners will
- Rearrange a quadratic equation so that one side equals zero.
- Solve factorised quadratic equations using the null factor law.
- Use the quadratic formula when an equation does not factorise easily.
- Use the discriminant to find the number of real solutions.
Success looks like
- I can solve an equation such as (x − 7)(x + 2) = 0.
- I can factorise and solve an equation such as x^2 − 5x + 6 = 0.
- I can use the quadratic formula and round my solutions sensibly.
- I can say whether a quadratic has 0, 1 or 2 real solutions.
The big idea
A quadratic equation contains an x^2 term and can be written as ax^2 + bx + c = 0. Most quadratic equations have two solutions, which is why we cannot just "undo" the operations as we do for linear equations.
The key idea is the null factor law: if A × B = 0, then A = 0 or B = 0. So if (x − 2)(x − 3) = 0, either x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.
To use the null factor law, first rearrange the equation so that one side is 0, then factorise the other side. For example, x^2 = 3x + 10 becomes x^2 − 3x − 10 = 0, which factorises to (x − 5)(x + 2) = 0, so x = 5 or x = −2.
Never divide both sides by x. In x^2 = 4x, dividing by x gives only x = 4 and loses the solution x = 0. Instead, write x^2 − 4x = 0 and factorise to x(x − 4) = 0.
When an equation does not factorise neatly, use the quadratic formula: x = (−b ± sqrt(b^2 − 4ac)) / (2a). The ± sign means you calculate once with + and once with −.
The expression under the square root, b^2 − 4ac, is called the discriminant. If it is positive, there are two real solutions. If it is zero, there is exactly one. If it is negative, there are no real solutions, because you cannot take the square root of a negative number using real numbers.
Graphically, the solutions of ax^2 + bx + c = 0 are the x-intercepts of the parabola y = ax^2 + bx + c. A parabola can cross the x-axis twice, touch it once, or miss it altogether, matching the three cases of the discriminant.
Worked examples
Factorise and solve
Solve x^2 − 5x + 6 = 0.
- Find two numbers that multiply to 6 and add to −5: they are −2 and −3.
- Factorise: (x − 2)(x − 3) = 0.
- Null factor law: x − 2 = 0 or x − 3 = 0.
- Check x = 2: 4 − 10 + 6 = 0. Check x = 3: 9 − 15 + 6 = 0.
Answer: x = 2 or x = 3
A common factor
Solve x^2 = 4x.
- Rearrange so one side is zero: x^2 − 4x = 0.
- Take out the common factor x: x(x − 4) = 0.
- Null factor law: x = 0 or x − 4 = 0.
Answer: x = 0 or x = 4
Using the quadratic formula
Solve 2x^2 + 3x − 4 = 0, giving answers correct to two decimal places.
- Identify a = 2, b = 3, c = −4.
- Discriminant: b^2 − 4ac = 9 − 4 × 2 × (−4) = 9 + 32 = 41. It is positive, so there are two solutions.
- x = (−3 ± sqrt(41)) / 4, and sqrt(41) ≈ 6.4031.
- x ≈ (−3 + 6.4031) / 4 ≈ 0.85 or x ≈ (−3 − 6.4031) / 4 ≈ −2.35.
Answer: x ≈ 0.85 or x ≈ −2.35
Practice check
Have a go, then check your answers. Each answer comes with an explanation.
Answer guide for parents
Solve (x − 7)(x + 2) = 0.
x = 7 or x = −2 — x − 7 = 0 gives x = 7, and x + 2 = 0 gives x = −2.
Solve x^2 − 9 = 0.
x = 3 or x = −3 — x^2 − 9 = (x − 3)(x + 3) = 0, so x = 3 or x = −3. Both square to 9.
How many real solutions does x^2 + 4x + 5 = 0 have?
0 — The discriminant is 4^2 − 4 × 1 × 5 = 16 − 20 = −4, which is negative, so there are no real solutions.
Which is the correct factorisation of x^2 + x − 12?
(x + 4)(x − 3) — 4 × (−3) = −12 and 4 + (−3) = 1, so x^2 + x − 12 = (x + 4)(x − 3).
What is the best first step to solve x^2 = 3x + 10?
Rearrange to x^2 − 3x − 10 = 0 — To use the null factor law, one side must be zero. Rearranging gives x^2 − 3x − 10 = 0, which factorises to (x − 5)(x + 2) = 0.
Find the positive solution of x^2 − 3x − 10 = 0.
x = 5 — x^2 − 3x − 10 = (x − 5)(x + 2) = 0, so x = 5 or x = −2. The positive solution is 5.
Find the larger solution of x^2 − 11x + 24 = 0.
x = 8 — Two numbers that multiply to 24 and add to −11 are −3 and −8, so (x − 3)(x − 8) = 0. The solutions are 3 and 8; the larger is 8.
Calculate the discriminant of x^2 − 6x + 9 = 0.
0 — b^2 − 4ac = (−6)^2 − 4 × 1 × 9 = 36 − 36 = 0, so the equation has exactly one solution (x = 3).
Watch out for
- Applying the null factor law when the right-hand side is not zero, for example claiming (x − 1)(x − 2) = 6 means x − 1 = 6 or x − 2 = 6. Always rearrange to zero first.
- Dividing both sides by x and losing the solution x = 0.
- Taking only the positive square root, such as solving x^2 = 9 as x = 3 only. Remember that (−3)^2 = 9 too.
- Sign errors in the quadratic formula, especially with −b and with a negative c. Write a, b and c down, including their signs, before substituting.
Tips for parents
- Ask your learner to check every solution by substituting it back into the original equation. Both solutions should work.
- If factorising is a struggle, spend time on factorising practice alone before combining it with solving.
- Encourage a quick sketch of the parabola. Seeing two, one or no x-intercepts makes the discriminant easier to understand.
Go further
- A rectangle has a length 3 cm more than its width and an area of 40 cm^2. Write and solve a quadratic equation to find its dimensions (width 5 cm, length 8 cm).
- Throw a ball upwards and research how its height can be modelled by a quadratic. When would the height equal zero?
- Investigate how completing the square can be used to derive the quadratic formula.