Solve Using the Quadratic Formula 2x^2+8x-10=0

$2 x^{2} + 8 x - 10 = 0$

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 $a = 2$ , $b = 8$ , and $c = - 10$ into the quadratic formula and solve for $x$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (2 \cdot - 10)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Raise $8$ to the power of $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 2 \cdot - 10}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 10$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 8 \cdot - 10}}{2 \cdot 2}$

Multiply $- 8$ by $- 10$ .

$x = \frac{- 8 \pm \sqrt{64 + 80}}{2 \cdot 2}$

Add $64$ and $80$ .

$x = \frac{- 8 \pm \sqrt{144}}{2 \cdot 2}$

Rewrite $144$ as $12^{2}$ .

$x = \frac{- 8 \pm \sqrt{12^{2}}}{2 \cdot 2}$

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

$x = \frac{- 8 \pm 12}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 8 \pm 12}{4}$

Simplify $\frac{- 8 \pm 12}{4}$ .

$x = - 2 \pm 3$

Step 4

The final answer is the combination of both solutions.

$x = 1, - 5$