Solve Using the Quadratic Formula 2x^2+4x-30=0

$2 x^{2} + 4 x - 30 = 0$

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 $a = 2$ , $b = 4$ , and $c = - 30$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (2 \cdot - 30)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 2 \cdot - 30}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 30$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 8 \cdot - 30}}{2 \cdot 2}$

Multiply $- 8$ by $- 30$ .

$x = \frac{- 4 \pm \sqrt{16 + 240}}{2 \cdot 2}$

Add $16$ and $240$ .

$x = \frac{- 4 \pm \sqrt{256}}{2 \cdot 2}$

Rewrite $256$ as $16^{2}$ .

$x = \frac{- 4 \pm \sqrt{16^{2}}}{2 \cdot 2}$

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

$x = \frac{- 4 \pm 16}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 4 \pm 16}{4}$

Simplify $\frac{- 4 \pm 16}{4}$ .

$x = - 1 \pm 4$

Step 4

The final answer is the combination of both solutions.

$x = 3, - 5$