Solve Using the Quadratic Formula 5x^2+6x-63=0

$5 x^{2} + 6 x - 63 = 0$

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 $a = 5$ , $b = 6$ , and $c = - 63$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (5 \cdot - 63)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 5 \cdot - 63}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 63$ .

Multiply $- 4$ by $5$ .

$x = \frac{- 6 \pm \sqrt{36 - 20 \cdot - 63}}{2 \cdot 5}$

Multiply $- 20$ by $- 63$ .

$x = \frac{- 6 \pm \sqrt{36 + 1260}}{2 \cdot 5}$

Add $36$ and $1260$ .

$x = \frac{- 6 \pm \sqrt{1296}}{2 \cdot 5}$

Rewrite $1296$ as $36^{2}$ .

$x = \frac{- 6 \pm \sqrt{36^{2}}}{2 \cdot 5}$

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

$x = \frac{- 6 \pm 36}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 6 \pm 36}{10}$

Simplify $\frac{- 6 \pm 36}{10}$ .

$x = \frac{- 3 \pm 18}{5}$

Step 4

The final answer is the combination of both solutions.

$x = 3, - \frac{21}{5}$