Algebra

The Quadratic Formula: When to Use It

Four methods solve a quadratic. The discriminant tells you which one to reach for before you commit to any of them.

By the Gauth editorial teamHow this content is produced and checkedUpdated 10 September 2026

The idea

A quadratic equation asks a geometric question. The graph of $y = a x^{2} + b x + c$ is a parabola, and solving $a x^{2} + b x + c = 0$ means finding the x-values where that curve sits exactly on the horizontal axis. A parabola is symmetric, so it can meet a horizontal line twice, once, or not at all. That is why a quadratic has two solutions, one, or none — and why no method can ever produce three.

Every parabola is a stretched, shifted copy of $y = x^{2}$ . Completing the square makes that literal: it rewrites any quadratic as $a (x - h)^{2} + k$ , which is the basic parabola moved $h$ across and $k$ up. Once the equation is in that shape, solving it needs nothing clever, because a squared quantity is easy to undo.

The quadratic formula is what you get when you complete the square on the general equation rather than on a specific one. Somebody did the algebra once with letters instead of numbers, and the answer is a recipe that works forever. That is the whole story. If you remember this, the plus-or-minus stops being arbitrary and the discriminant stops being a piece of trivia.

When you need it

The recognition cue is a single squared variable and nothing of higher degree. After you expand and collect, the highest power of $x$ is two, and the coefficient of $x^{2}$ is not zero. That is the entire entry condition.

These are the phrasings that mean “solve a quadratic”:

Find the roots, the zeros, or the x-intercepts of a function — all three name the same thing.
Where does the projectile hit the ground? Height is quadratic in time, so set $h (t) = 0$ .
A rectangle problem where one dimension is described in terms of the other and you are given the area. Multiplying two linear expressions produces a quadratic.
Where do a line and a parabola intersect, or where do two parabolas cross? Setting the expressions equal collapses to one quadratic.
A rational equation such as $\frac{x}{2} - \frac{3}{x} = 1$ . Clearing the denominator raises the degree, and a quadratic drops out.

Two near-misses are worth naming. An equation like $x^{4} - 5 x^{2} + 4 = 0$ is not a quadratic, but it is quadratic in $x^{2}$ , so the same machinery applies after you substitute $u = x^{2}$ . And if the $x^{2}$ terms cancel during simplification, you have a linear equation with exactly one solution — do not force the formula onto it, because $a = 0$ makes the denominator zero.

The method

Before the recipe, here is where it comes from. The derivation is six lines and worth doing once, because it explains every feature of the result.

a x^{2} + b x + c = 0, a \neq = 0

Divide through by $a$ so the squared term is bare, then move the constant across:

x^{2} + \frac{b}{a} x = - \frac{c}{a}

The left side is the start of a perfect square. Since $(x + \frac{b}{2 a})^{2} = x^{2} + \frac{b}{a} x + \frac{b ^{2}}{4 a ^{2}}$ , add $\frac{b ^{2}}{4 a ^{2}}$ to both sides:

(x + \frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}} - \frac{c}{a} = \frac{b ^{2} - 4 a c}{4 a ^{2}}

Take the square root of both sides. The right-hand side has denominator $4 a^{2}$ , whose root is $2∣ a ∣$ , but the plus-or-minus already covers both signs, so $2 a$ is safe to write:

x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{2 a} ⟹ x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Notice what fell out. The discriminant $b^{2} - 4 a c$ appeared because it is the numerator left over after combining $\frac{b ^{2}}{4 a ^{2}} - \frac{c}{a}$ . The plus-or-minus appeared because square roots come in pairs. Nothing was invented.

Working a specific equation, the steps are:

Get everything on one side. Expand brackets, clear fractions, and collect terms until the equation reads $a x^{2} + b x + c = 0$ . Coefficients read off any other arrangement are meaningless.
Name a, b and c with their signs attached. In $3 x^{2} - 5 x - 2 = 0$ the values are $a = 3$ , $b = - 5$ , $c = - 2$ . The minus signs belong to the coefficients.
Evaluate the discriminant on its own line.Doing this first costs one line and buys you the answer to “how many roots” and “will this factor” before you commit to any arithmetic.
Substitute, bracketing the negative b. Writing $- (- 5)$ explicitly rather than trying to hold the double negative in your head prevents the most frequent slip on this page.
Simplify the surd, then cancel. Reduce $60$ to $215$ before you look for common factors, or you will miss the cancellation.
Substitute both roots back. Fifteen seconds of checking catches sign errors that no amount of re-reading will.

Three worked examples

The first is the textbook case. The second produces irrational roots that need genuine simplification. The third is deliberately awkward — it arrives as a rational equation and has to be dragged into standard form before the formula is even legal.

Worked example

2 x^{2} - 7 x + 3 = 0

1
Already in standard form, so read off the coefficients directly.
$a = 2, b = - 7, c = 3$
2
Evaluate the discriminant first. It is 25, a perfect square, so the roots will be rational and this equation would also have factored.
$b^{2} - 4 a c = (- 7)^{2} - 4 (2) (3) = 49 - 24 = 25$
3
Substitute. Bracket the negative b so the double negative becomes a clean positive seven.
$x = \frac{- ( - 7 ) \pm 25}{2 ( 2 )} = \frac{7 \pm 5}{4}$
4
Split the plus-or-minus into the two separate roots.
$x = \frac{7 + 5}{4} = 3 or x = \frac{7 - 5}{4} = \frac{1}{2}$
5
Check both. Substituting 3 gives 18 − 21 + 3 = 0, and substituting one half gives 0.5 − 3.5 + 3 = 0.
$2 (3)^{2} - 7 (3) + 3 = 0, 2 (\frac{1}{2})^{2} - 7 (\frac{1}{2}) + 3 = 0$

Answer

x = 3 or x = \frac{1}{2}

Worked example

3 x^{2} + 6 x - 2 = 0

1
Coefficients, with the sign of c attached.
$a = 3, b = 6, c = - 2$
2
The discriminant is positive but not a perfect square, so expect two irrational roots. There is no integer factorisation to look for.
$b^{2} - 4 a c = 6^{2} - 4 (3) (- 2) = 36 + 24 = 60$
3
Substitute, then simplify the surd before doing anything else. Sixty has a factor of four.
$x = \frac{- 6 \pm 60}{6} = \frac{- 6 \pm 2 15}{6}$
4
Every term in the numerator and the denominator shares a factor of two. Cancel it — and cancel it from all three parts, not just the two you notice first.
$x = \frac{2 ( - 3 \pm 15 )}{6} = \frac{- 3 \pm 15}{3}$
5
Sanity-check numerically. The square root of fifteen is about 3.873, giving roots near 0.291 and −2.291, and their sum is −2, which matches −b/a = −2.
$x \approx 0.291 or x \approx - 2.291$

Answer

x = \frac{- 3 + 15}{3} or x = \frac{- 3 - 15}{3}

Worked example

\frac{x}{2} - \frac{3}{x} = 1

1
This is not a quadratic yet, and x = 0 is barred from the outset because it would make the second fraction undefined. Note that restriction now, before it can be forgotten.
$x \neq = 0$
2
Multiply every term by 2x, the lowest common denominator. Multiply the term on the right as well — forgetting it is the classic error here.
$2 x \cdot \frac{x}{2} - 2 x \cdot \frac{3}{x} = 2 x \cdot 1 ⟹ x^{2} - 6 = 2 x$
3
Now collect on one side to reach standard form.
$x^{2} - 2 x - 6 = 0, a = 1, b = - 2, c = - 6$
4
Discriminant: positive, not a perfect square, so irrational roots again.
$b^{2} - 4 a c = (- 2)^{2} - 4 (1) (- 6) = 4 + 24 = 28$
5
Substitute and simplify. Twenty-eight is four times seven, so the radical reduces and a factor of two cancels throughout.
$x = \frac{2 \pm 28}{2} = \frac{2 \pm 2 7}{2} = 1 \pm 7$
6
Both roots are non-zero, so the restriction is satisfied and neither is extraneous. Numerically, 1 + √7 ≈ 3.646 gives 1.823 − 0.823 = 1, as required.
$1 + 7 \approx 3.646, 1 - 7 \approx - 1.646$

Answer

x = 1 + 7 or x = 1 - 7

Pick your method before you start

Reaching for the formula every time is a defensible habit, but it is often the slowest route. Walk this once for the equation in front of you and the choice usually takes under ten seconds.

Which method should you use?

Does the equation contain a linear x term once it is set equal to zero?

The discriminant also has a visual meaning. Drag the coefficients below and watch the parabola: when the curve cuts the axis twice the discriminant is positive, when it just kisses the axis the discriminant is zero, and when it floats clear of the axis the discriminant has gone negative. The algebra and the picture are the same fact.

Move a, b and c and watch the roots

y = 1x² + 0x + -2

Solve a quadratic problem now

First solution free

Where people go wrong

Reading coefficients off an unarranged equation

Given $2 x^{2} + 5 x = 3$ , it is tempting to set $c = 3$ . It is not: moving the three across makes $c = - 3$ , and the discriminant changes from $25 - 24 = 1$ to $25 + 24 = 49$ . One is a near-miss of a double root, the other has roots at $\frac{1}{2}$ and $- 3$ . Rearrange first, every time.

Dividing only part of the numerator by 2a

Writing $x = - b \pm \frac{b ^{2} - 4 a c}{2 a}$ leaves $- b$ outside the fraction. The entire numerator sits over $2 a$ . A quick check catches it: the two roots must sum to $- b / a$ , because the plus and minus parts cancel when you add them. If your roots do not sum correctly, this is usually why.

Cancelling before factoring the numerator

In $\frac{- 6 \pm 2 15}{6}$ , the six in the denominator cannot be cancelled against the six in the numerator alone. Factor the numerator into $2 (- 3 \pm 15)$ first, then cancel the common factor of two. Cancelling a term rather than a factor is the single most costly habit in algebra, and it shows up here more than anywhere else.

Squaring a negative b incorrectly

With $b = - 7$ , the term $b^{2}$ is $(- 7)^{2} = 49$ , not $- 49$ . The exponent binds to the whole coefficient, brackets or no brackets, because $b$ is the number $- 7$ . This is the same precedence question that makes $- 3^{2} = - 9$ while $(- 3)^{2} = 9$ , which is unpicked in full in the guide to how order of operations really works.

Practice

Work these on paper before typing. For equations with two roots, list the smaller one first and separate them with a comma.

Four quadratics, four different situations

Answers are checked here — nothing is sent anywhere.

1
$x^{2} - 5 x + 6 = 0 (both roots, smallest first)$
2
$2 x^{2} + 3 x - 2 = 0 (both roots, smallest first)$
3
$How many real roots does 2 x^{2} - 4 x + 5 = 0 have?$
4
$4 x^{2} - 12 x + 9 = 0 (give the single root)$

Where this goes next

Quadratics are the first place where a method genuinely has to be chosen rather than followed, which is why the same equation appears again in the four factoring patterns worth recognising — a perfect-square discriminant is precisely the signal that one of those patterns is hiding in your equation. Beyond a single equation, pairing a quadratic with a line is a system, and the choice of substitution, elimination or graphing follows a similar decision rule. If the arithmetic rather than the algebra is what keeps going wrong, the quadratic formula calculator runs all four methods on the same equation so you can see where your working diverged.

Frequently asked questions

Does the quadratic formula work on every quadratic?

Yes, provided the equation is written as ax^2 + bx + c = 0 with a not equal to zero. That is the only requirement. Factoring, the square-root method, and completing the square each need something extra to be true about the coefficients; the formula needs nothing. That reliability is the reason to learn it, even though it is often not the fastest route.

What does the discriminant actually tell me?

b^2 - 4ac is the quantity under the square root, so its sign decides how many real roots exist. Positive gives two distinct real roots, zero gives one repeated root where the parabola touches the axis, negative gives no real roots and two complex conjugates. If it is a positive perfect square the roots are rational, which is exactly the condition for the quadratic to factor over the integers.

Why is there a plus-or-minus in the formula?

Because the last step before isolating x is taking a square root, and a positive number has two square roots. Completing the square turns the equation into (x + b/2a)^2 = (b^2 - 4ac)/4a^2, and undoing that square admits both the positive and the negative root. The two signs are what produce the two solutions.

Can I just use a calculator for this?

For a numerical answer, yes. But most marks in an exam are for method, not the final number, and a calculator will not show you that the discriminant was a perfect square and the whole thing factored in five seconds. Use a tool to check your working rather than to replace it.

What do I do when the discriminant is negative?

Write the roots in the form (-b +/- i*sqrt(4ac - b^2)) / 2a. Note the sign flip inside the radical: you factor out the negative one as i, so the quantity under the root becomes 4ac - b^2, which is positive. If the question is set in the real numbers, the correct answer is simply that there are no real solutions, and it is worth saying why: the parabola sits entirely above or entirely below the x-axis.

Keep going

Algebra solverEvery step shown for equations, inequalities, and systems, not just quadratics.Quadratic formula calculatorSolve one equation four ways side by side and compare the working.Factoring quadraticsThe faster route when the discriminant is a perfect square.Systems of equationsWhat to do when a quadratic meets a second equation.