Solve Using the Quadratic Formula 12x-4-9x^2=0

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

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

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

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 $- 4$ 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 $- 12$ .

$x = \frac{- 6 \cdot 2}{- 18}$

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.

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

The final answer is the combination of both solutions.

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