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

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

Add $8 x^{2}$ to both sides of the equation.

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 8$ , $b = - 2$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (8 \cdot 3)}}{2 \cdot 8}$

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 8 \cdot 3}}{2 \cdot 8}$

Multiply $- 4$ by $8$ .

$x = \frac{2 \pm \sqrt{4 - 32 \cdot 3}}{2 \cdot 8}$

Multiply $- 32$ by $3$ .

$x = \frac{2 \pm \sqrt{4 - 96}}{2 \cdot 8}$

Subtract $96$ from $4$ .

$x = \frac{2 \pm \sqrt{- 92}}{2 \cdot 8}$

Rewrite $- 92$ as $- 1 (92)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 92}}{2 \cdot 8}$

Rewrite $\sqrt{- 1 (92)}$ as $\sqrt{- 1} \cdot \sqrt{92}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{92}}{2 \cdot 8}$

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

$x = \frac{2 \pm i \cdot \sqrt{92}}{2 \cdot 8}$

Rewrite $92$ as $2^{2} \cdot 23$ .

Factor $4$ out of $92$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (23)}}{2 \cdot 8}$

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

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 23}}{2 \cdot 8}$

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{23})}{2 \cdot 8}$

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{23}}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{2 \pm 2 i \sqrt{23}}{16}$

Simplify $\frac{2 \pm 2 i \sqrt{23}}{16}$ .

$x = \frac{1 \pm i \sqrt{23}}{8}$