The Quadratic Formula

Learning goals

Derive the formula by completing the square in general
Divide the whole numerator, both $-b$ and the radical, by $2a$
Read $D = b^2 - 4ac$ to sort the roots
Write standard form and read $a$ , $b$ , $c$ with their signs
Simplify the radical, then reduce the fraction
Accept complex roots when the discriminant is negative

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, so we earn it step by step. Every move below is one you already made in the last lesson on specific numbers; the only new thing is that the numbers are now letters.

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

Start from the general quadratic equation, with the single restriction that $a \ne 0$ . 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. A square root carries two signs, so attach a $\pm$ , and split the root over numerator and denominator:

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

The denominator simplified because $\sqrt{4a^2} = 2a$ . (Strictly the root is $2\lvert a\rvert$ , but the $\pm$ in front already delivers both signs, so writing $2a$ throws nothing away.) 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 $a$ , $b$ , 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.

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 number is called the discriminant, and it alone decides what kind of roots a quadratic has. The reason is that the discriminant is the thing you are about to take the square root of. 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

Notice you can name the type of roots from $D$ alone, without ever finishing the solve. Computing the discriminant first is a quick way to know what you are heading toward.

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 is the exact border between real and complex roots, and it always signals that the quadratic is a perfect square, 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.

That single procedure, put in standard form and substitute, now handles every quadratic you will meet. The next lesson turns it loose on 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.
Losing the sign of $-b$ . The numerator begins with $-b$ , the opposite of $b$ . If $b = -6$ then $-b = +6$ , not $-6$ . Substitute $b$ with its sign and let $-b$ flip it.
Computing $b^2$ as a negative. Squaring always gives a nonnegative result, so $b = -5$ makes $b^2 = 25$ , never $-25$ . The minus sign disappears under the square.
Getting the sign of $-4ac$ wrong. When $a$ and $c$ have opposite signs, $-4ac$ turns positive and adds to $b^2$ . 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 $-$ . Keeping only the plus throws away half the answer.
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 $\pm$ produces the two roots at once.
The discriminant $D = b^2 - 4ac$ sorts the roots: $D > 0$ gives two distinct real roots, $D = 0$ gives one repeated real root $x = -\tfrac{b}{2a}$ , and $D < 0$ gives a complex conjugate pair.
To apply it, write the equation in standard form, read off $a$ , $b$ , and $c$ with their signs, then substitute. Simplify the radical (for example $\sqrt{24} = 2\sqrt{6}$ ), then reduce the fraction by any factor shared with the denominator.
A negative discriminant is no longer a dead end: since $\sqrt{-1} = i$ , the roots are complex, so every quadratic has two roots, real or complex.

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)

Free Response Questions (FRQ)