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Quadratic equations: from intuition to the formula

By Daniel Whitman · Last updated April 2026 · 16 min read

A quadratic equation is any equation that, after you tidy it up, looks like $ax^2 + bx + c = 0$ with $a \ne 0$.

That is the textbook definition, and it is correct, but it is also the thing that makes quadratics feel arbitrary the first time you meet them. Why this one shape? Why do we care so much about an equation involving $x^2$?

The honest answer is that quadratics are the simplest equations whose graphs bend. A linear equation like $3x - 5 = 0$ has a straight-line graph, and the question “where does it cross zero?” has at most one answer. The moment you let $x^2$ into the picture, the graph becomes a parabola — a curve that turns around. That curve can cross the x-axis twice, touch it once, or miss it entirely. The whole story of solving quadratic equations is the story of telling those three cases apart, and finding the crossings when they exist.

This guide is the lesson I would give in a single class. We’ll cover the three solving methods (factoring, completing the square, and the formula), why the discriminant predicts the answer count before you ever solve, and the small handful of mistakes that account for almost every wrong answer I have ever marked.

What “quadratic” actually means

The word quadratic comes from the Latin quadratus, meaning “square.” It refers not to the shape of the parabola but to the $x^2$ term — literally, “x squared.” Any equation in which the highest power of the unknown is two is quadratic. So $x^2 = 9$ is quadratic; so is $3x^2 + 2x = 1$; and so is the slightly disguised $x(x - 4) = 5$. The first thing you do with any of these is rearrange them into the standard form $ax^2 + bx + c = 0$ by moving everything to one side.

That standard form matters because it is the form every method below is designed to work on. If you skip it — if you try to solve $x^2 = 5x - 6$ by setting both sides equal to each other “in your head” — you will, eventually, drop a sign and lose a root.

Three ways to solve $x^2 - 5x + 6 = 0$

Let’s solve the same equation three different ways. The point is not that you should do all three on a test (you shouldn’t); the point is that each method shows you something different about how the equation fits together.

Method 1 — factoring

We want two numbers that multiply to $6$ and add to $-5$. The pair $-2$ and $-3$ works: their product is $6$ and their sum is $-5$. So we can write

$$x^2 - 5x + 6 \;=\; (x - 2)(x - 3) \;=\; 0.$$

Now we use the one fact about real numbers that makes factoring useful: a product is zero only when one of the factors is zero. So either $x - 2 = 0$, giving $x = 2$, or $x - 3 = 0$, giving $x = 3$.

Factoring is the fastest method when it works, but it only works cleanly when the roots are nice numbers — usually integers or simple fractions. For $2x^2 + 7x - 4 = 0$ factoring is still possible but already painful; for $x^2 - x - 1 = 0$ (whose roots are the golden ratio and its conjugate) it is impossible without already knowing the answer.

Method 2 — completing the square

This method always works, and it is also the proof of where the quadratic formula comes from. The idea is to rewrite the left side as a perfect square plus a constant.

Start by isolating the x terms: $x^2 - 5x = -6$. Then add to both sides whatever turns the left into a square. The general rule is: take half the coefficient of $x$, square it, and add it. Half of $-5$ is $-\tfrac{5}{2}$, and $\left(-\tfrac{5}{2}\right)^2 = \tfrac{25}{4}$. So:

$$x^2 - 5x + \tfrac{25}{4} \;=\; -6 + \tfrac{25}{4} \;=\; \tfrac{1}{4}$$ $$\left(x - \tfrac{5}{2}\right)^2 \;=\; \tfrac{1}{4}$$ $$x - \tfrac{5}{2} \;=\; \pm \tfrac{1}{2}$$ $$x \;=\; \tfrac{5}{2} \pm \tfrac{1}{2} \;=\; 2 \text{ or } 3.$$

Notice the $\pm$ step. That is the place where the second root sneaks in. When students forget the $\pm$, they get one root and lose the other.

Method 3 — the quadratic formula

The quadratic formula is what completing the square gives you when you carry it out in full generality on $ax^2 + bx + c = 0$. The result is

$$x \;=\; \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$

Substituting $a = 1$, $b = -5$, $c = 6$:

$$x \;=\; \frac{5 \pm \sqrt{25 - 24}}{2} \;=\; \frac{5 \pm 1}{2} \;=\; 3 \text{ or } 2.$$

The formula is the brute-force option. It always works, it does not care whether the roots are nice, and on an exam it is almost always the right tool when the factoring is not immediate.

The discriminant: what is under the square root

The expression $b^2 - 4ac$ inside the square root has a name — the discriminant, usually written $\Delta$. Its sign alone tells you how many real roots the equation has, without computing them.

Δ > 0 Δ = 0 Δ < 0
The three things the discriminant can tell you, before you solve.

A useful exam habit: before you reach for the formula, compute $\Delta$. If $\Delta < 0$ the question is asking for complex roots (or there are none, depending on your syllabus). If $\Delta = 0$ you can stop and write the answer as $-b/(2a)$. If $\Delta$ is a perfect square, the equation factors over the rationals and you might want to factor instead. A thirty-second discriminant check often saves three minutes of arithmetic.

The mistakes I see most often

After eight years of marking quadratics, the same handful of errors come up again and again. They are worth recognising because they are predictable.

1. Forgetting to set the equation to zero

Students see $x^2 + 3x = 4$ and try to factor the left side. The factoring technique only works when the right side is zero, because we need the “a product is zero implies a factor is zero” rule. Move the $4$ over first: $x^2 + 3x - 4 = 0$, then factor as $(x + 4)(x - 1) = 0$, giving $x = -4$ or $x = 1$.

2. Dividing both sides by $x$

Faced with $x^2 = 4x$, the temptation is to cancel an $x$ and write $x = 4$. That loses the root $x = 0$. Whenever you cancel a variable, you implicitly assume it is non-zero, and you lose the case where it is. Bring everything to one side and factor instead: $x^2 - 4x = 0$, $x(x - 4) = 0$, so $x = 0$ or $x = 4$.

3. Sign errors inside the discriminant

$b^2$ is always non-negative, even when $b$ is negative: $(-5)^2 = 25$, not $-25$. And $4ac$ takes its sign from the product of $a$ and $c$ — if $c$ is negative, then $-4ac$ becomes positive in the discriminant. Most “no real roots” mistakes I see in homework come from miscomputing this single quantity, not from a deeper conceptual problem.

4. Dropping the $\pm$ after the square root

A square root of a positive number has two values, and the $\pm$ in the formula is the only thing that captures both. The most common version of this mistake is writing only the $+$ branch and missing the second root entirely.

5. Treating $ax^2 + b = 0$ as a special case that “needs” the formula

You can use the formula here (with $c$ where you expect, in the right slot) and it will give the right answer, but it is faster to recognise that when there is no $x$ term you can just isolate $x^2$ and take a square root: $4x^2 - 9 = 0 \Rightarrow x^2 = \tfrac{9}{4} \Rightarrow x = \pm\tfrac{3}{2}$.

Beyond two real roots

The formula does not stop working when the discriminant is negative; it just gives roots that involve $i$, the imaginary unit defined by $i^2 = -1$. For example, $x^2 + 2x + 5 = 0$ has $\Delta = 4 - 20 = -16$, and the formula gives

$$x \;=\; \frac{-2 \pm \sqrt{-16}}{2} \;=\; -1 \pm 2i.$$

These are called complex conjugate roots, and they always come in pairs because of the $\pm$. Whether your syllabus expects you to write them down depends on the course: GCSE does not, A-Level does, and so does most first-year university maths.

Vieta’s formulas

Two facts about quadratic roots fall out of the formula directly and are sometimes useful as a shortcut. They are called Vieta’s formulas:

For $x^2 - 5x + 6 = 0$, the sum should be $5$ and the product should be $6$ — and indeed the roots $2$ and $3$ sum to $5$ and multiply to $6$. Vieta’s formulas are most useful when a problem asks you for the sum or product of roots without asking for the roots themselves, or when you are constructing a quadratic with given roots.

Choosing a method on a test

Reach for factoring first if the coefficients are small integers and the constant term has obvious factor pairs. It is the fastest method when it works.

Reach for completing the square when a problem asks you for the vertex of a parabola, or for the maximum or minimum value of a quadratic expression, rather than its roots. The completed form $(x - h)^2 + k$ tells you the vertex $(h, k)$ directly, with no further calculation.

Reach for the quadratic formula for everything else — especially when the coefficients are not small integers, or when factoring is not producing nice numbers within about fifteen seconds. It is the only method that is genuinely fool-proof, and on a timed exam being fool-proof is worth more than being fast.

Where this comes up later

Once you can solve quadratics quickly, three things become easier. First, intersection problems — “where does this line cross this parabola?” — reduce to a quadratic equation. Second, optimisation problems in basic calculus often produce a quadratic to solve at the end. And third, simple physics problems involving uniform acceleration (range, maximum height, time of flight) almost always come down to setting a quadratic equal to zero. The technique is general; the practice is specific.

References and further reading