Solve Using the Quadratic Formula n^2-14n=-45

$n^{2} - 14 n = - 45$

Step 1

Add $45$ to both sides of the equation.

$n^{2} - 14 n + 45 = 0$

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 $a = 1$ , $b = - 14$ , and $c = 45$ into the quadratic formula and solve for $n$ .

$\frac{14 \pm \sqrt{{(- 14)}^{2} - 4 \cdot (1 \cdot 45)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 14$ to the power of $2$ .

$n = \frac{14 \pm \sqrt{196 - 4 \cdot 1 \cdot 45}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 45$ .

Multiply $- 4$ by $1$ .

$n = \frac{14 \pm \sqrt{196 - 4 \cdot 45}}{2 \cdot 1}$

Multiply $- 4$ by $45$ .

$n = \frac{14 \pm \sqrt{196 - 180}}{2 \cdot 1}$

Subtract $180$ from $196$ .

$n = \frac{14 \pm \sqrt{16}}{2 \cdot 1}$

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

$n = \frac{14 \pm \sqrt{4^{2}}}{2 \cdot 1}$

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

$n = \frac{14 \pm 4}{2 \cdot 1}$

Multiply $2$ by $1$ .

$n = \frac{14 \pm 4}{2}$

Simplify $\frac{14 \pm 4}{2}$ .

$n = 7 \pm 2$

Step 5

The final answer is the combination of both solutions.

$n = 9, 5$