Solve Using the Quadratic Formula x^2+14x+45=0

$x^{2} + 14 x + 45 = 0$

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

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $45$ .

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

Subtract $180$ from $196$ .

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

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

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

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

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

Multiply $2$ by $1$ .

$x = \frac{- 14 \pm 4}{2}$

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

$x = - 7 \pm 2$

Step 4

The final answer is the combination of both solutions.

$x = - 5, - 9$