Quadratic equations — how to solve them step by step

A quadratic equation is an equation of the form ax² + bx + c = 0 where a is not zero. It is one of the most important equations at GCSE, and it appears in the algebra papers, in problems about motion, in physics and in economics. In this guide you will find the precise definition, the quadratic formula, an explanation of the discriminant and its three cases, three complete worked examples, the factorising method, the link to the parabola, real-life uses and the common mistakes to avoid.

A quadratic equation is an algebraic equation of the form ax² + bx + c = 0, where a, b and c are constants and a is not zero. The variable x appears squared, hence the name 'quadratic' (from the Latin for square). It is one of the most basic and most common equations in secondary maths, and it appears in almost every part of the GCSE — from algebra, through coordinate geometry, to word problems and trigonometry.

This guide takes you through everything you need to know: the definition, the quadratic formula, the discriminant, three detailed examples, the other solving methods, the link to the parabola and uses in real life.

What is a quadratic equation?

A quadratic equation is an equation in which the highest power of the unknown is 2. Its general form is:

ax² + bx + c = 0

where:

• **a** — the coefficient of x² (must be non-zero, otherwise the equation is linear, not quadratic)
• **b** — the coefficient of x (may be zero)
• **c** — the constant term, a number with no x (may be zero)

Examples of quadratic equations:

• 2x² + 3x − 5 = 0 (here a = 2, b = 3, c = −5)
• x² − 9 = 0 (here a = 1, b = 0, c = −9)
• 4x² + 8x = 0 (here a = 4, b = 8, c = 0)

What does it mean to 'solve a quadratic equation'? Solving means finding all the values of x that satisfy the equation — that is, values which, when substituted for x, give zero. These values are called the roots of the equation, or its solutions.

A quadratic equation can have two solutions, one (repeated) solution, or no real solution — it depends on the equation. Later we will see how to know in advance which case you are facing.

The quadratic formula — the central tool

The most important tool for solving quadratic equations is the quadratic formula:

x = (−b ± √(b² − 4ac)) / 2a

This formula always works, on every quadratic equation, without exception. It is worth knowing by heart — since the 2017 reforms the GCSE papers no longer give it on a formula sheet, and it is examined on the Higher tier.

How do you use it? Three simple steps:

1. **Identify a, b and c** from the equation (after rearranging it into the standard form, with everything on the left and zero on the right).
2. **Substitute** the values into the formula.
3. **Calculate the two solutions** — one with the + sign and one with the − sign.

The expression under the square root, b² − 4ac, is called the discriminant (written with the Greek letter delta: Δ). It is the little secret of the quadratic equation — it tells us in advance how many solutions there will be.

If you want to see how the formula works on a specific equation with a guided solution, the interactive MathsUK solver carries out every step in real time — you can enter any equation and get a step-by-step breakdown.

The discriminant — three cases

The discriminant is the expression under the square root in the quadratic formula:

Δ = b² − 4ac

The sign of Δ decides how many solutions the equation has. There are three possible cases:

Case 1: Δ > 0 → two different solutions

If the discriminant is positive, its square root is a positive real number, and thanks to the ± sign we get two different solutions — one larger and one smaller. Graphically, this means the parabola crosses the x-axis at two points.

Case 2: Δ = 0 → one (repeated) solution

If the discriminant equals zero, its square root is zero, and the formula reduces to x = −b/2a. There is only one solution, called a 'repeated root' because algebraically it counts twice. Graphically, the parabola touches the x-axis at exactly one point.

Case 3: Δ < 0 → no real solution

If the discriminant is negative, you cannot take a square root within the real numbers. So the equation has no solution in the set of real numbers. Graphically, the parabola does not cross the x-axis at all.

An important tip: before you start solving, work out the discriminant. If it is negative — you are done, there is no solution. It saves time and prevents mistakes.

Three worked examples, step by step

Example 1: x² − 5x + 6 = 0 (two solutions)

Step 1 — identify the coefficients: a = 1, b = −5, c = 6

Step 2 — calculate the discriminant: Δ = b² − 4ac = (−5)² − 4·1·6 = 25 − 24 = 1

The discriminant is positive (Δ = 1 > 0), so we expect two solutions.

Step 3 — substitute into the quadratic formula: x = (−(−5) ± √1) / (2·1) = (5 ± 1) / 2

Step 4 — calculate the two solutions:

• x₁ = (5 + 1) / 2 = 6/2 = **3**
• x₂ = (5 − 1) / 2 = 4/2 = **2**

Step 5 — check (important!): substitute x = 3 into the equation: 3² − 5·3 + 6 = 9 − 15 + 6 = 0 ✓. Substitute x = 2: 2² − 5·2 + 6 = 4 − 10 + 6 = 0 ✓

Example 2: x² − 4x + 4 = 0 (a repeated root)

Step 1: a = 1, b = −4, c = 4. Step 2: Δ = (−4)² − 4·1·4 = 16 − 16 = 0. The discriminant is zero — one repeated solution. Step 3: x = (4 ± 0) / 2 = 4/2 = 2. Step 4 — check: 2² − 4·2 + 4 = 4 − 8 + 4 = 0 ✓

The only solution is x = 2 (a repeated root).

Example 3: x² + 2x + 5 = 0 (no solution)

Step 1: a = 1, b = 2, c = 5. Step 2: Δ = 2² − 4·1·5 = 4 − 20 = −16. The discriminant is negative.

Conclusion: the equation has no real solution. The parabola y = x² + 2x + 5 does not cross the x-axis. You can confirm this visually in the function grapher and see the parabola floating above the axis.

Factorising — the elegant method

The quadratic formula always works, but sometimes there is a quicker route: factorising. The idea is to write the equation as a product of two linear factors that equals zero:

(x − r₁)(x − r₂) = 0

If a product equals zero, one of the factors must be zero. So the solutions are x = r₁ or x = r₂.

How do you factorise? Look for two numbers whose product is c and whose sum is b (when a = 1).

Example: x² − 5x + 6 = 0. Look for two numbers whose product is 6 and whose sum is −5. They are −2 and −3. The factorisation: (x − 2)(x − 3) = 0. The solutions: x = 2 or x = 3 — exactly as with the quadratic formula, but much faster. Factorising is the method examined on the Foundation tier; the formula and completing the square are Higher.

Other methods

• **Completing the square:** the technique that explains where the quadratic formula comes from. Also used at GCSE Higher to find the turning point of the parabola.
• **Taking out a common factor of x:** when c = 0, it is enough to factor out x. For example: x² + 3x = 0 → x(x + 3) = 0 → x = 0 or x = −3.
• **Direct solution:** when b = 0, you can isolate x². For example: x² − 9 = 0 → x² = 9 → x = ±3.

When is each method best?

• **Factorising** — when you can immediately see two numbers that fit.
• **Common factor or direct solution** — when b = 0 or c = 0.
• **The quadratic formula** — in every other case. The safe default.

The link to the parabola

Every quadratic equation is linked to a quadratic function of the form y = ax² + bx + c, whose graph is a parabola. The roots of the equation are exactly the points where the parabola crosses the x-axis — that is, the x-values for which y = 0.

• If **Δ > 0** — the parabola crosses the x-axis at two points.
• If **Δ = 0** — the parabola touches the x-axis at one point.
• If **Δ < 0** — the whole parabola lies above or below the x-axis.

Understanding this link is essential for GCSE. To practise it, go to the GCSE algebra practice and work on questions that combine an algebraic solution with reading the graph.

Uses in real life

Quadratic equations are not just a school exercise — they appear wherever there is a squared relationship between quantities.

Motion problems (free fall and vertical throws)

The height of an object thrown upwards is given by the formula h = −5t² + v₀t + h₀, where h is the height in metres, t the time in seconds, v₀ the initial speed and h₀ the initial height. If you want to know when the object hits the ground, you solve the equation h = 0 — a classic quadratic.

Area and dimensions

A typical question: 'The length of a rectangle is 3 metres more than its width, and its area is 40 m². What are the dimensions of the rectangle?' If the width is x, the area is x(x + 3) = 40, giving x² + 3x − 40 = 0.

Economics and business

Many profit functions are quadratic, and finding the maximum profit or the 'break-even' points (zero profit) leads to a quadratic equation.

Speed and time problems

Almost every word problem that combines different speeds in different directions ends in a quadratic equation. To practise this type of problem, you can use the mock papers, which have a wide bank of them.

Common mistakes

The following mistakes are the main reason for losing marks in exams:

1. **Forgetting the ±** — pupils substitute into the formula and forget to calculate both solutions. Always separate into x₁ and x₂ explicitly.
2. **A sign error in b** — if b = −5, then −b = +5, not −5. Use brackets when substituting: (−b) becomes (−(−5)) = 5.
3. **Miscalculating Δ** — forgetting that b² is always positive (even when b is negative), or an error in multiplying 4ac.
4. **Not checking the solution** — a minute of checking (substituting into the original equation) saves lost marks.
5. **Confusing a quadratic equation with a quadratic inequality** — an inequality also needs an analysis of where the sign changes, not just the roots.
6. **Forgetting to rearrange into standard form** — if the equation is given as 3x² + 2 = 5x, everything must first be moved to one side.
7. **Hidden division by zero** — if you cancel a factor, make sure you have not lost the solution where that factor equals zero.

For simpler (linear) equations you can use the linear equation solver to go back over the basics.

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