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

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

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 $c = - 9$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 4 \cdot - 9$ .

Multiply $- 4$ by $4$ .

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

Multiply $- 16$ by $- 9$ .

$x = \frac{- 9 \pm \sqrt{81 + 144}}{2 \cdot 4}$

Add $81$ and $144$ .

$x = \frac{- 9 \pm \sqrt{225}}{2 \cdot 4}$

Rewrite $225$ as $15^{2}$ .

$x = \frac{- 9 \pm \sqrt{15^{2}}}{2 \cdot 4}$

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

$x = \frac{- 9 \pm 15}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 9 \pm 15}{8}$

Step 4

The final answer is the combination of both solutions.

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