Solve Using the Quadratic Formula 2x^2+10x-48=0

$2 x^{2} + 10 x - 48 = 0$

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 $a = 2$ , $b = 10$ , and $c = - 48$ into the quadratic formula and solve for $x$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (2 \cdot - 48)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Raise $10$ to the power of $2$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 2 \cdot - 48}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 48$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 10 \pm \sqrt{100 - 8 \cdot - 48}}{2 \cdot 2}$

Multiply $- 8$ by $- 48$ .

$x = \frac{- 10 \pm \sqrt{100 + 384}}{2 \cdot 2}$

Add $100$ and $384$ .

$x = \frac{- 10 \pm \sqrt{484}}{2 \cdot 2}$

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

$x = \frac{- 10 \pm \sqrt{22^{2}}}{2 \cdot 2}$

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

$x = \frac{- 10 \pm 22}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 10 \pm 22}{4}$

Simplify $\frac{- 10 \pm 22}{4}$ .

$x = \frac{- 5 \pm 11}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 3, - 8$