The Quadratic Formula

Learning goals

Derive the quadratic formula by completing the square on $ax^2 + bx + c = 0$
Write a quadratic in standard form and read off $a$ , $b$ , $c$ with their signs
Apply the formula, simplifying radicals and reducing the fraction to find the roots
Use $D = b^2 - 4ac$ to predict real, repeated, or complex roots before solving

Deriving the formula by completing the square

The quadratic formula is not a rule to accept on faith. It is exactly what completing the square gives when you feed it the general equation. Every move below is one you already made in the last lesson on specific numbers, for example $a = 2$ , $b = 5$ , $c = -3$ in $2x^2 + 5x - 3 = 0$ . The only new thing is that now those numbers stay as letters all the way through.

Deriving the quadratic formula from $ax^2 + bx + c = 0$ #

Start from the general quadratic equation, where $a$ , $b$ , and $c$ are real numbers with $a \ne 0$ . If $a$ were $0$ there would be no $x^2$ term and the equation would not be quadratic at all:

ax^2 + bx + c = 0.

Completing the square needs the leading coefficient to be $1$ , so divide every term by $a$ . Because $a \ne 0$ , dividing by it is allowed:

x^2 + \frac{b}{a}x + \frac{c}{a} = 0.

Move the constant term to the right side, so the left holds only the two terms that carry $x$ :

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

Now complete the square on the left. Half of the coefficient of $x$ is $\dfrac{b}{2a}$ , and squaring it gives $\dfrac{b^2}{4a^2}$ . Add that constant to both sides, which keeps the equation balanced:

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

The left side is now a perfect square. Since half of $\tfrac{b}{a}$ is $\tfrac{b}{2a}$ , it factors as

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

Combine the right side into one fraction. The common denominator is $4a^2$ , and rewriting $\dfrac{c}{a} = \dfrac{4ac}{4a^2}$ lets the two pieces subtract cleanly:

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{4ac}{4a^2} = \frac{b^2 - 4ac}{4a^2}.

Take the square root of both sides. Squaring erases a sign, so undoing it creates two possible equations, one for each sign the squared side could have had; attach a $\pm$ to capture both, and split the root over numerator and denominator. ( $\sqrt{\phantom{x}}$ itself always names one fixed value, never two: $\sqrt{9}$ names $3$ and not $-3$ , exactly as established in Imaginary Numbers. The two signs come from the $\pm$ , not from the radical.) Since $4a^2$ is a perfect square, its root simplifies the denominator to $2a$ (the note after this derivation explains the one subtlety in that step):

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

Finally subtract $\dfrac{b}{2a}$ from both sides to isolate $x$ :

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

Both terms on the right already share the denominator $2a$ , so they merge into a single fraction, and that fraction is the quadratic formula:

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

Because the derivation never assumed anything about the real numbers $a$ , $b$ , and $c$ beyond $a \ne 0$ , the formula holds for every quadratic equation. Whatever completing the square would have found, the formula finds too, with none of the repeated setup.

Check your understanding

In the derivation, why must every term be divided by $a$ before completing the square?

A It puts the equation into the familiar pattern

x^2 + (\text{something})x

B Dividing by

a

removes the need for the

\pm

later C It makes the equation shorter to write D It only works when

a

is negative

Check your understanding

In the derivation, why is $\left(\dfrac{b}{2a}\right)^2$ , and not $\dfrac{b}{2a}$ or $\dfrac{b}{a}$ , the constant added to both sides to complete the square?

A Completing the square always adds the square of half the coefficient of

x

, and here that coefficient is

\dfrac{b}{a}

, so half of it, squared, is

\left(\dfrac{b}{2a}\right)^2

B Squaring is needed to cancel the fraction

\dfrac{b}{a}

C The formula only works when the added constant is itself squared D Any constant added to both sides keeps the equation balanced, so the exact value does not matter

What the formula says

Here is the result on its own, worth memorizing because you will use it constantly:

ax^2 + bx + c = 0 \quad\Longrightarrow\quad x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

The single $\pm$ packs both solutions into one line. Read with a plus it gives one root; read with a minus it gives the other. The one part people misread is the denominator: the $2a$ divides the entire numerator, both the $-b$ and the $\sqrt{b^2 - 4ac}$ together, not just one piece of it.

The anatomy of the quadratic formula. Everything in the numerator, the negative b term and the square-root term together, sits over 2a, so the whole top is divided by 2a. The quantity b squared minus 4ac beneath the root is the discriminant, and its sign decides whether the roots are real, repeated, or complex.

To use the formula you first need the three numbers $a$ , $b$ , and $c$ . They are the coefficients when the equation is written in standard form, with everything on one side and zero on the other:

ax^2 + bx + c = 0.

Here $a$ is the coefficient of $x^2$ , $b$ is the coefficient of $x$ , and $c$ is the constant term. Read them with their signs. In $3x^2 - 5x + 2 = 0$ the coefficients are $a = 3$ , $b = -5$ , and $c = 2$ ; the minus sign belongs to $b$ . Getting a sign wrong here is the fastest way to a wrong answer, so it is worth a moment of care before you substitute.

The discriminant

Look again at the quantity under the square root:

D = b^2 - 4ac.

This is called the discriminant, and it alone decides what kind of roots a quadratic has. (If you compare it with the quantity $R$ from the last lesson’s derivation, for $a = 1$ this one equals $4R$ . Multiplying by a positive number never flips a sign, so the same three cases below still hold.) There are exactly three cases, and each comes straight from what a square root does to a positive number, to zero, and to a negative number.

If $D > 0$ , then $\sqrt{D}$ is a nonzero real number, and the $\pm$ pulls it once above and once below $-b$ , giving two different real roots.

If $D = 0$ , then $\sqrt{D} = 0$ , and adding or subtracting $0$ makes no difference, so the two roots fall together into one repeated real root, $x = -\dfrac{b}{2a}$ .

If $D < 0$ , then $\sqrt{D}$ is the square root of a negative number, which is imaginary. Writing $\sqrt{D} = i\sqrt{\lvert D\rvert}$ turns the answer into two complex roots, a conjugate pair $-\dfrac{b}{2a} \pm \dfrac{\sqrt{\lvert D\rvert}}{2a}\,i$ . Before this chapter that case simply had no answer; now that $i$ is available, it does.

Discriminant $D = b^2 - 4ac$	The square root $\sqrt{D}$	Roots of the quadratic
$D > 0$	a nonzero real number	two distinct real roots
$D = 0$	zero	one repeated real root
$D < 0$	imaginary	two complex conjugate roots

You can name the type of roots from $D$ alone, without ever finishing the solve.

Check your understanding

Without solving, what does the discriminant tell you about the roots of $2x^2 - 3x + 5 = 0$ ?

A two complex conjugate roots B two distinct real roots C one repeated real root D no roots at all

Applying the formula

With the formula, the discriminant, and the imaginary unit in hand, one procedure now solves any quadratic. Put the equation in standard form, read off $a$ , $b$ , and $c$ with their signs, substitute, and simplify.

Worked example 1 Solve $2x^2 + 5x - 3 = 0$

The equation is already in standard form, so read off $a = 2$ , $b = 5$ , and $c = -3$ , keeping the sign on $c$ . Substitute into the formula, watching the double negative in $-4ac$ :

x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 24}}{4}.

The term $-4(2)(-3) = +24$ is positive because $c$ was negative, so the discriminant is $25 + 24 = 49$ , a perfect square:

x = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}.

Now split the $\pm$ into its two cases:

x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}, \qquad x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3.

The two roots are $x = \tfrac{1}{2}$ and $x = -3$ . When the discriminant is a perfect square, the roots come out rational, exactly the case that ordinary factoring could also have caught.

Worked example 2 Solve $3x^2 - 6x + 1 = 0$

Read off $a = 3$ , $b = -6$ , and $c = 1$ . Because $b = -6$ , the term $-b$ is $+6$ , and $b^2 = (-6)^2 = 36$ is positive; a common slip is to write $-36$ here. Substitute:

x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(1)}}{2(3)} = \frac{6 \pm \sqrt{36 - 12}}{6} = \frac{6 \pm \sqrt{24}}{6}.

The discriminant $24$ is not a perfect square, so the roots are irrational. Simplify the radical using $24 = 4 \times 6$ and $\sqrt{4} = 2$ :

x = \frac{6 \pm 2\sqrt{6}}{6}.

Every term in the fraction shares a factor of $2$ , so divide the numerator and denominator by $2$ to reduce:

x = \frac{2\left(3 \pm \sqrt{6}\right)}{6} = \frac{3 \pm \sqrt{6}}{3}.

The roots are $x = \dfrac{3 + \sqrt{6}}{3}$ and $x = \dfrac{3 - \sqrt{6}}{3}$ , both real and irrational. Reduce only by a factor common to $-b$ , the radical, and the denominator; here that shared factor is $2$ .

Check your understanding

Solve $2x^2 + 4x - 1 = 0$ with the quadratic formula.

x = -1 \pm 2\sqrt{6}

x = \dfrac{-2 \pm \sqrt{6}}{2}

x = \dfrac{-4 \pm \sqrt{6}}{2}

x = \dfrac{2 \pm \sqrt{6}}{2}

The formula does not stop at real roots. When the discriminant is negative, the same substitution runs straight into a square root of a negative number, which the imaginary unit $i$ handles.

Worked example 3 Solve $x^2 + 4x + 13 = 0$ (complex roots)

Read off $a = 1$ , $b = 4$ , and $c = 13$ , and substitute:

x = \frac{-4 \pm \sqrt{4^2 - 4(1)(13)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2}.

The discriminant is $-36$ , negative, so the roots are complex. Rewrite the root of the negative with $i$ , using $\sqrt{-36} = i\sqrt{36} = 6i$ :

x = \frac{-4 \pm 6i}{2}.

Divide both parts of the numerator by $2$ :

x = \frac{-4}{2} \pm \frac{6i}{2} = -2 \pm 3i.

The roots are the conjugate pair $x = -2 + 3i$ and $x = -2 - 3i$ . Whenever the discriminant is negative, the two roots always arrive as a conjugate pair like this, matching in real part and opposite in imaginary part.

Worked example 4 Solve $4x^2 - 12x + 9 = 0$ (a repeated root)

Read off $a = 4$ , $b = -12$ , and $c = 9$ . Substitute, being careful that $-b = 12$ and $b^2 = (-12)^2 = 144$ :

x = \frac{12 \pm \sqrt{(-12)^2 - 4(4)(9)}}{2(4)} = \frac{12 \pm \sqrt{144 - 144}}{8} = \frac{12 \pm \sqrt{0}}{8}.

The discriminant is $0$ , so $\sqrt{0} = 0$ and the $\pm$ adds nothing. The two roots collapse into one:

x = \frac{12}{8} = \frac{3}{2}.

There is a single repeated root, $x = \tfrac{3}{2}$ . A zero discriminant always signals that the quadratic factors as $a$ times a squared linear expression, $a(x - r)^2$ , so the same root repeats twice; here $4x^2 - 12x + 9 = (2x - 3)^2$ .

Check your understanding

Solve $x^2 - 6x + 10 = 0$ with the quadratic formula.

x = -3 \pm i

x = 3 \pm 2i

x = 3 \pm i\sqrt{10}

x = 3 \pm i

A quadratic does not always arrive in standard form. When it does not, move every term to one side first, so that the other side is $0$ , and only then read off $a$ , $b$ , and $c$ .

Worked example 5 Solve $x^2 = 3x + 10$ by first reaching standard form

As written, the equation is not set to $0$ , so you cannot read the coefficients yet. Subtract $3x$ and $10$ from both sides to gather everything on the left:

x^2 - 3x - 10 = 0.

Now it is in standard form with $a = 1$ , $b = -3$ , and $c = -10$ . Substitute, noting that $-4ac = -4(1)(-10) = +40$ :

x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2}.

The discriminant is $49$ , so

x = \frac{3 \pm 7}{2} \quad\Rightarrow\quad x = 5 \ \text{ or }\ x = -2.

Reaching standard form first is what made $a$ , $b$ , and $c$ readable. Skipping that step, and reading coefficients off $x^2 = 3x + 10$ directly, is a reliable way to get the signs wrong.

Check your understanding

Solve $x^2 + x = 12$ with the quadratic formula. (Reach standard form first.)

x = 3

x = -4

x = 4

x = -3

x = 6

x = -8

x = \dfrac{-1 \pm i\sqrt{47}}{2}

That single procedure, put in standard form and substitute, now handles every quadratic you will meet. The next lesson applies it to real-world situations, where an unknown length, time, or price is whatever value makes a quadratic equal zero.

Common mistakes

Dividing only part of the numerator by $2a$ . The whole numerator $-b \pm \sqrt{b^2 - 4ac}$ sits over $2a$ . Writing $-b \pm \dfrac{\sqrt{b^2 - 4ac}}{2a}$ , with only the radical divided, drops the division on the $-b$ and gives a wrong root.
Mishandling signs when substituting. The numerator starts with $-b$ , the opposite of $b$ : if $b = -6$ then $-b = +6$ , not $-6$ . Squaring always gives a nonnegative result, so $b^2$ is never negative. And $-4ac$ turns positive whenever $a$ and $c$ have opposite signs; in $2x^2 + 5x - 3 = 0$ the discriminant is $25 - 4(2)(-3) = 25 + 24 = 49$ , not $25 - 24$ .
Dropping the $\pm$ . The formula produces two roots, one with $+$ and one with $-$ (the same value twice when $D = 0$ ). Keeping only the plus misses a root whenever $D \ne 0$ .
Calling a negative discriminant “no solution.” With $i$ available, $D < 0$ gives a complex conjugate pair, not nothing. Write $\sqrt{-36} = 6i$ and finish.
Substituting before standard form. Read $a$ , $b$ , and $c$ only after the equation is set equal to $0$ . Coefficients pulled from $x^2 = 3x + 10$ carry the wrong signs.

Key takeaways

Completing the square on the general equation $ax^2 + bx + c = 0$ (with $a \ne 0$ ) gives the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , which solves every quadratic. The entire numerator, both $-b$ and $\sqrt{b^2 - 4ac}$ , is divided by $2a$ .
The discriminant $D = b^2 - 4ac$ sorts the roots before you finish solving. $D > 0$ gives two distinct real roots, $D = 0$ gives one repeated real root $x = -\tfrac{b}{2a}$ , and $D < 0$ gives a complex conjugate pair, available now that $\sqrt{-1} = i$ .
To apply the formula, write the equation in standard form, read off $a$ , $b$ , and $c$ with their signs, then substitute. Watch the sign flips on $-b$ , $b^2$ , and $-4ac$ .
After substituting, simplify the radical and reduce the fraction by any factor common to both numerator terms and $2a$ .

The Quadratic Formula

Learning goals

Deriving the formula by completing the square

What the formula says

The discriminant

Applying the formula

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice