Find the Roots (Zeros) 12=-3x^2+16x

$12 = - 3 x^{2} + 16 x$

Step 1

Rewrite the equation as $- 3 x^{2} + 16 x = 12$ .

$- 3 x^{2} + 16 x = 12$

Step 2

Subtract $12$ from both sides of the equation.

$- 3 x^{2} + 16 x - 12 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = - 3$ , $b = 16$ , and $c = - 12$ into the quadratic formula and solve for $x$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (- 3 \cdot - 12)}}{2 \cdot - 3}$

Step 5

Simplify.

Simplify the numerator.

Raise $16$ to the power of $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot - 3 \cdot - 12}}{2 \cdot - 3}$

Multiply $- 4 \cdot - 3 \cdot - 12$ .

Multiply $- 4$ by $- 3$ .

$x = \frac{- 16 \pm \sqrt{256 + 12 \cdot - 12}}{2 \cdot - 3}$

Multiply $12$ by $- 12$ .

$x = \frac{- 16 \pm \sqrt{256 - 144}}{2 \cdot - 3}$

Subtract $144$ from $256$ .

$x = \frac{- 16 \pm \sqrt{112}}{2 \cdot - 3}$

Rewrite $112$ as $4^{2} \cdot 7$ .

Factor $16$ out of $112$ .

$x = \frac{- 16 \pm \sqrt{16 (7)}}{2 \cdot - 3}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 16 \pm \sqrt{4^{2} \cdot 7}}{2 \cdot - 3}$

Pull terms out from under the radical.

$x = \frac{- 16 \pm 4 \sqrt{7}}{2 \cdot - 3}$

Multiply $2$ by $- 3$ .

$x = \frac{- 16 \pm 4 \sqrt{7}}{- 6}$

Simplify $\frac{- 16 \pm 4 \sqrt{7}}{- 6}$ .

$x = \frac{8 \pm 2 \sqrt{7}}{3}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{8 \pm 2 \sqrt{7}}{3}$

Decimal Form:

$x = 4.43050087 \dots, 0.90283245 \dots$

Step 7