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

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

Step 1

Add $9$ to both sides of the equation.

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

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

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

$x = \frac{- 12 \pm \sqrt{144 - 16 \cdot 9}}{2 \cdot 4}$

Multiply $- 16$ by $9$ .

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

Subtract $144$ from $144$ .

$x = \frac{- 12 \pm \sqrt{0}}{2 \cdot 4}$

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

$x = \frac{- 12 \pm \sqrt{0^{2}}}{2 \cdot 4}$

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

$x = \frac{- 12 \pm 0}{2 \cdot 4}$

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

$x = \frac{- 12}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 12}{8}$

Cancel the common factor of $- 12$ and $8$ .

Factor $4$ out of $- 12$ .

$x = \frac{4 (- 3)}{8}$

Cancel the common factors.

Factor $4$ out of $8$ .

$x = \frac{4 \cdot - 3}{4 \cdot 2}$

Cancel the common factor.

$x = \frac{4 \cdot - 3}{4 \cdot 2}$

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Step 5

The final answer is the combination of both solutions.

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