Solve Using the Quadratic Formula 5x^2+14x-48=0

$5 x^{2} + 14 x - 48 = 0$

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 $a = 5$ , $b = 14$ , and $c = - 48$ into the quadratic formula and solve for $x$ .

$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot (5 \cdot - 48)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 14 \pm \sqrt{196 - 4 \cdot 5 \cdot - 48}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 48$ .

Multiply $- 4$ by $5$ .

$x = \frac{- 14 \pm \sqrt{196 - 20 \cdot - 48}}{2 \cdot 5}$

Multiply $- 20$ by $- 48$ .

$x = \frac{- 14 \pm \sqrt{196 + 960}}{2 \cdot 5}$

Add $196$ and $960$ .

$x = \frac{- 14 \pm \sqrt{1156}}{2 \cdot 5}$

Rewrite $1156$ as $34^{2}$ .

$x = \frac{- 14 \pm \sqrt{34^{2}}}{2 \cdot 5}$

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

$x = \frac{- 14 \pm 34}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 14 \pm 34}{10}$

Simplify $\frac{- 14 \pm 34}{10}$ .

$x = \frac{- 7 \pm 17}{5}$

Step 4

The final answer is the combination of both solutions.

$x = 2, - \frac{24}{5}$