How to Solve Quadratic Equations: Factorising, Formula and Completing the Square

What Is a Quadratic Equation?

A quadratic equation is any equation where the highest power of the unknown is 2, so it contains an $x^2$ term and nothing higher. The standard form is $ax^2 + bx + c = 0$, where $a$, $b$ and $c$ are numbers and $a$ is not zero. Solving the equation means finding every value of $x$ that makes it true, and those values are called the roots or solutions.

Quadratics behave differently from the linear equations you meet first. A linear equation like $3x + 2 = 11$ has exactly one solution. A quadratic usually has two, because squaring destroys sign information: both $3^2$ and $(-3)^2$ equal 9, so an equation built around $x^2$ generally has two ways of being satisfied. Sometimes the two roots coincide (a repeated root), and sometimes there are no real solutions at all — the discriminant section below tells you which case you are in before you do any solving.

This topic runs right through both tiers. Foundation papers ask you to solve quadratics that factorise; Higher papers add the quadratic formula, completing the square and equations that appear inside other topics such as simultaneous equations. The quadratics topic page breaks down exactly what each tier expects; this guide teaches the three methods themselves, then shows you how to pick between them under exam conditions.

Before You Solve Anything: Rearrange to Equal Zero

Every method in this guide starts from the same place: one side of the equation must be zero. If a question hands you $x^2 + 3x = 10$, your first line of working is $x^2 + 3x - 10 = 0$. Skipping this step is the single most common way to go wrong on this topic, because the logic of factorising depends on it entirely.

Here is why. Factorising turns the equation into the form $(x + p)(x + q) = 0$, and the only reason you can then split it into two mini-equations is a fact about zero: if two things multiply to give zero, at least one of them must be zero. That fact is unique to zero. If $(x + p)(x + q) = 10$, you can conclude nothing — the brackets could be 2 and 5, or 20 and 0.5, or anything else. Rearranging to zero is not a stylistic habit, it is the step that makes the whole method valid.

The other skill to have sharp before you start is expanding and factorising itself, because solving by factorising is exactly that skill run in reverse. If you cannot yet reliably turn $x^2 - 5x + 6$ into $(x - 2)(x - 3)$, work through expanding and factorising brackets first — ten minutes there will save you many marks here.

Method 1: Factorising

When the quadratic factorises neatly, this is the fastest method and the one to reach for first. For $x^2 + bx + c$, you are hunting for two numbers that multiply to give $c$ and add to give $b$. Those two numbers go straight into the brackets, and then each bracket is set equal to zero in turn.

Signs do most of the damage on these questions, so read them off systematically. If $c$ is positive, both numbers share a sign, and $b$ tells you which one. If $c$ is negative, the two numbers have opposite signs. In the example below, $c = +6$ and $b = -5$, so you need two negative numbers: $-2$ and $-3$.

Higher papers also use quadratics where $a > 1$, such as $2x^2 + 7x + 3 = 0$. These factorise as $(2x + 1)(x + 3) = 0$, giving $x = -\frac{1}{2}$ or $x = -3$ — notice that the root from the bracket $2x + 1$ is $-\frac{1}{2}$, not $-1$. Always finish by solving each bracket properly rather than just flipping the sign of the number you see.

Method 2: The Quadratic Formula

The quadratic formula solves any quadratic, including the ones that refuse to factorise. For $ax^2 + bx + c = 0$:

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

You must know this from memory — it is not on the formulae sheet any exam board provides. Write it down at the start of your answer, then write a line stating your values of $a$, $b$ and $c$ before substituting. That one line is worth real marks and catches most sign errors before they happen, because the commonest mistake with the formula is mishandling a negative $b$ or a negative $c$.

Two calculator habits protect your accuracy. First, evaluate the discriminant $b^2 - 4ac$ on its own before touching the rest of the formula, and write it down. Second, keep the exact value on your calculator and only round at the final line. A question that says "give your answers correct to 2 decimal places" is telling you two things: use the formula, because the quadratic will not factorise, and do not round early.

Method 3: Completing the Square

Completing the square rewrites $x^2 + bx + c$ in the form $(x + p)^2 + q$. The recipe: halve the coefficient of $x$ to get $p$, then subtract $p^2$ to correct for the extra term the squared bracket creates, and finally bring in the constant. So $x^2 + 6x$ becomes $(x + 3)^2 - 9$, because expanding $(x + 3)^2$ gives $x^2 + 6x + 9$ and the $-9$ cancels the 9 you never had.

Once the quadratic is in completed-square form, solving it takes three short moves: isolate the squared bracket, square-root both sides remembering the $\pm$, and rearrange. Forgetting the $\pm$ costs you exactly half the answer — one root instead of two — and it is the mistake examiners see most on this method.

Completing the square earns its place for two reasons. It produces exact answers in surd form, which is what "give your answer in the form $a \pm \sqrt{b}$" is asking for. And the completed-square form reads off the graph's turning point directly, which the sketching section below uses. When $a > 1$, factor $a$ out of the $x$ terms first: $2x^2 + 8x + 3 = 2(x^2 + 4x) + 3 = 2(x + 2)^2 - 8 + 3$.

The Discriminant: Knowing How Many Roots Before You Solve (Higher)

The expression under the square root in the quadratic formula, $b^2 - 4ac$, is called the discriminant, and it answers one question: how many real roots does this quadratic have? You can answer it before doing any solving at all.

The intuition comes straight from the formula. The $\pm$ in $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ is what creates two answers. If the discriminant is positive, the square root is a genuine positive number, so adding it and subtracting it give two different roots. If the discriminant is exactly zero, you are adding and subtracting nothing, so both versions collapse into the single root $x = -\frac{b}{2a}$. And if the discriminant is negative, you would need the square root of a negative number, which no real number provides — so there are no real roots.

Try it on the examples above: $x^2 - 5x + 6$ has discriminant $25 - 24 = 1 > 0$, two roots. $x^2 - 6x + 9$ has discriminant $36 - 36 = 0$, one repeated root ($x = 3$). $x^2 + 4x + 5$ has discriminant $16 - 20 = -4 < 0$, no real roots. Higher papers ask this both directly ("show that the equation has no real roots") and in reverse ("find the values of $k$ for which $x^2 + kx + 9 = 0$ has equal roots" — set $k^2 - 36 = 0$, so $k = \pm 6$).

What the Roots Look Like on a Graph

Every quadratic equation has a picture behind it. The graph of $y = ax^2 + bx + c$ is a parabola — a symmetric U-shape opening upwards when $a > 0$ and downwards when $a < 0$ — and solving $ax^2 + bx + c = 0$ asks where that curve has height zero. In other words, the roots are exactly the $x$-intercepts of the graph.

This picture makes the discriminant visual. Two distinct roots: the parabola cuts the $x$-axis in two places. A repeated root: the parabola just touches the axis at its turning point, like $y = (x - 3)^2$. No real roots: the whole curve floats above or below the axis and never reaches it.

Each algebraic form of the quadratic hands you a different feature of the sketch, which is why sketching questions reward knowing all three methods. The factorised form gives the $x$-intercepts. The constant $c$ gives the $y$-intercept, since $x = 0$ makes $y = c$. And the completed-square form $(x + p)^2 + q$ gives the turning point at $(-p, q)$, because a squared bracket is smallest — zero — when $x = -p$. For $y = x^2 + 6x + 7 = (x + 3)^2 - 2$, that means a minimum at $(-3, -2)$, a $y$-intercept at 7, and roots at $-3 \pm \sqrt{2}$: three facts, one clean sketch.

Choosing the Right Method Under Exam Conditions

In the exam you rarely get to dither, so use the question's own wording as your guide. "Solve" with integer-looking coefficients on a non-calculator paper: try factorising first, and give it no more than thirty seconds — if the two numbers will not appear, the question probably wants another method. "Give your answers correct to 2 decimal places" or "to 3 significant figures": use the formula, because rounded answers are the signature of a quadratic that does not factorise. "Give exact answers", "in the form $a \pm \sqrt{b}$", or anything mentioning the turning point: complete the square.

A worked comparison: faced with $x^2 + 6x + 7 = 0$, you look for two numbers multiplying to 7 and adding to 6. The only factor pair of 7 is 1 and 7, which adds to 8 — so factorising is dead, and you move on without wasting time. That quick test, multiply-and-add on the factor pairs of $c$, is how you decide in seconds rather than minutes.

Whatever method you choose, show the marked steps: the factorisation itself, the stated values of $a$, $b$, $c$ and the discriminant, or the completed-square form. Most quadratic questions carry 3 to 4 marks and the majority are method marks, so a visible strategy protects you even when the arithmetic slips. Exam Ladder's mock exams are graded against real exam-board grade boundaries, so you can see exactly what those part-marks do to a grade — and its practice engine draws on over 280,000 generated questions across every GCSE Maths topic when you want more quadratics than any textbook holds.