Find the Roots (Zeros) 2x^2-8x+12=-3x

$2 x^{2} - 8 x + 12 = - 3 x$

Move all terms containing $x$ to the left side of the equation.

Add $3 x$ to both sides of the equation.

$2 x^{2} - 8 x + 12 + 3 x = 0$

Add $- 8 x$ and $3 x$ .

$2 x^{2} - 5 x + 12 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = - 5$ , and $c = 12$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (2 \cdot 12)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{5 \pm \sqrt{25 - 8 \cdot 12}}{2 \cdot 2}$

Multiply $- 8$ by $12$ .

$x = \frac{5 \pm \sqrt{25 - 96}}{2 \cdot 2}$

Subtract $96$ from $25$ .

$x = \frac{5 \pm \sqrt{- 71}}{2 \cdot 2}$

Rewrite $- 71$ as $- 1 (71)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 71}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (71)}$ as $\sqrt{- 1} \cdot \sqrt{71}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{71}}{2 \cdot 2}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{5 \pm i \sqrt{71}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{5 \pm i \sqrt{71}}{4}$