Solve Using the Quadratic Formula x^2+2x-120=0

$x^{2} + 2 x - 120 = 0$

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 $a = 1$ , $b = 2$ , and $c = - 120$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 120)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 120}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 120$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 120}}{2 \cdot 1}$

Multiply $- 4$ by $- 120$ .

$x = \frac{- 2 \pm \sqrt{4 + 480}}{2 \cdot 1}$

Add $4$ and $480$ .

$x = \frac{- 2 \pm \sqrt{484}}{2 \cdot 1}$

Rewrite $484$ as $22^{2}$ .

$x = \frac{- 2 \pm \sqrt{22^{2}}}{2 \cdot 1}$

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

$x = \frac{- 2 \pm 22}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 22}{2}$

Simplify $\frac{- 2 \pm 22}{2}$ .

$x = - 1 \pm 11$

Step 4

The final answer is the combination of both solutions.

$x = 10, - 12$