Solve Using the Quadratic Formula 9x^2+12x=-4

$9 x^{2} + 12 x = - 4$

Step 1

Add $4$ to both sides of the equation.

$9 x^{2} + 12 x + 4 = 0$

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 $a = 9$ , $b = 12$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (9 \cdot 4)}}{2 \cdot 9}$

Step 4

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 9 \cdot 4}}{2 \cdot 9}$

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

Multiply $- 4$ by $9$ .

$x = \frac{- 12 \pm \sqrt{144 - 36 \cdot 4}}{2 \cdot 9}$

Multiply $- 36$ by $4$ .

$x = \frac{- 12 \pm \sqrt{144 - 144}}{2 \cdot 9}$

Subtract $144$ from $144$ .

$x = \frac{- 12 \pm \sqrt{0}}{2 \cdot 9}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{- 12 \pm \sqrt{0^{2}}}{2 \cdot 9}$

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

$x = \frac{- 12 \pm 0}{2 \cdot 9}$

$- 12$ plus or minus $0$ is $- 12$ .

$x = \frac{- 12}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{- 12}{18}$

Cancel the common factor of $- 12$ and $18$ .

Factor $6$ out of $- 12$ .

$x = \frac{6 (- 2)}{18}$

Cancel the common factors.

Factor $6$ out of $18$ .

$x = \frac{6 \cdot - 2}{6 \cdot 3}$

Cancel the common factor.

$x = \frac{6 \cdot - 2}{6 \cdot 3}$

Rewrite the expression.

$x = \frac{- 2}{3}$

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

Step 5

The final answer is the combination of both solutions.

$x = - \frac{2}{3}$ Double roots