Solve Using the Quadratic Formula 5x^2-47x=156

$5 x^{2} - 47 x = 156$

Step 1

Subtract $156$ from both sides of the equation.

$5 x^{2} - 47 x - 156 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 5$ , $b = - 47$ , and $c = - 156$ into the quadratic formula and solve for $x$ .

$\frac{47 \pm \sqrt{{(- 47)}^{2} - 4 \cdot (5 \cdot - 156)}}{2 \cdot 5}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 47$ to the power of $2$ .

$x = \frac{47 \pm \sqrt{2209 - 4 \cdot 5 \cdot - 156}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 156$ .

Multiply $- 4$ by $5$ .

$x = \frac{47 \pm \sqrt{2209 - 20 \cdot - 156}}{2 \cdot 5}$

Multiply $- 20$ by $- 156$ .

$x = \frac{47 \pm \sqrt{2209 + 3120}}{2 \cdot 5}$

Add $2209$ and $3120$ .

$x = \frac{47 \pm \sqrt{5329}}{2 \cdot 5}$

Rewrite $5329$ as $73^{2}$ .

$x = \frac{47 \pm \sqrt{73^{2}}}{2 \cdot 5}$

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

$x = \frac{47 \pm 73}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{47 \pm 73}{10}$

Step 5

The final answer is the combination of both solutions.

$x = 12, - \frac{13}{5}$