Solve Using the Quadratic Formula x^2-12x+40=0

$x^{2} - 12 x + 40 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 1$ , $b = - 12$ , and $c = 40$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (1 \cdot 40)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 12$ to the power of $2$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 1 \cdot 40}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 40$ .

Multiply $- 4$ by $1$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 40}}{2 \cdot 1}$

Multiply $- 4$ by $40$ .

$x = \frac{12 \pm \sqrt{144 - 160}}{2 \cdot 1}$

Subtract $160$ from $144$ .

$x = \frac{12 \pm \sqrt{- 16}}{2 \cdot 1}$

Rewrite $- 16$ as $- 1 (16)$ .

$x = \frac{12 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \frac{12 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{12 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{12 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{12 \pm i \cdot 4}{2 \cdot 1}$

Move $4$ to the left of $i$ .

$x = \frac{12 \pm 4 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{12 \pm 4 i}{2}$

Simplify $\frac{12 \pm 4 i}{2}$ .

$x = 6 \pm 2 i$