Solve Using the Quadratic Formula 4x^2+9x-63=0

$4 x^{2} + 9 x - 63 = 0$

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 $a = 4$ , $b = 9$ , and $c = - 63$ into the quadratic formula and solve for $x$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (4 \cdot - 63)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $9$ to the power of $2$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 4 \cdot - 63}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$x = \frac{- 9 \pm \sqrt{81 - 16 \cdot - 63}}{2 \cdot 4}$

Multiply $- 16$ by $- 63$ .

$x = \frac{- 9 \pm \sqrt{81 + 1008}}{2 \cdot 4}$

Add $81$ and $1008$ .

$x = \frac{- 9 \pm \sqrt{1089}}{2 \cdot 4}$

Rewrite $1089$ as $33^{2}$ .

$x = \frac{- 9 \pm \sqrt{33^{2}}}{2 \cdot 4}$

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

$x = \frac{- 9 \pm 33}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 9 \pm 33}{8}$

Step 4

The final answer is the combination of both solutions.

$x = 3, - \frac{21}{4}$