Find All Complex Solutions n^2-13n=-40

$n^{2} - 13 n = - 40$

Step 1

Add $40$ to both sides of the equation.

$n^{2} - 13 n + 40 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = - 13$ , and $c = 40$ into the quadratic formula and solve for $n$ .

$\frac{13 \pm \sqrt{{(- 13)}^{2} - 4 \cdot (1 \cdot 40)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 13$ to the power of $2$ .

$n = \frac{13 \pm \sqrt{169 - 4 \cdot 1 \cdot 40}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$n = \frac{13 \pm \sqrt{169 - 4 \cdot 40}}{2 \cdot 1}$

Multiply $- 4$ by $40$ .

$n = \frac{13 \pm \sqrt{169 - 160}}{2 \cdot 1}$

Subtract $160$ from $169$ .

$n = \frac{13 \pm \sqrt{9}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

$n = \frac{13 \pm \sqrt{3^{2}}}{2 \cdot 1}$

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

$n = \frac{13 \pm 3}{2 \cdot 1}$

Multiply $2$ by $1$ .

$n = \frac{13 \pm 3}{2}$

Step 5

The final answer is the combination of both solutions.

$n = 8, 5$