Find the Roots (Zeros) 9=-2x^2+3x

$9 = - 2 x^{2} + 3 x$

Rewrite the equation as $- 2 x^{2} + 3 x = 9$ .

$- 2 x^{2} + 3 x = 9$

Subtract $9$ from both sides of the equation.

$- 2 x^{2} + 3 x - 9 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = - 2$ , $b = 3$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (- 2 \cdot - 9)}}{2 \cdot - 2}$

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 2 \cdot - 9}}{2 \cdot - 2}$

Multiply $- 4$ by $- 2$ .

$x = \frac{- 3 \pm \sqrt{9 + 8 \cdot - 9}}{2 \cdot - 2}$

Multiply $8$ by $- 9$ .

$x = \frac{- 3 \pm \sqrt{9 - 72}}{2 \cdot - 2}$

Subtract $72$ from $9$ .

$x = \frac{- 3 \pm \sqrt{- 63}}{2 \cdot - 2}$

Rewrite $- 63$ as $- 1 (63)$ .

$x = \frac{- 3 \pm \sqrt{- 1 \cdot 63}}{2 \cdot - 2}$

Rewrite $\sqrt{- 1 (63)}$ as $\sqrt{- 1} \cdot \sqrt{63}$ .

$x = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{63}}{2 \cdot - 2}$

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

$x = \frac{- 3 \pm i \cdot \sqrt{63}}{2 \cdot - 2}$

Rewrite $63$ as $3^{2} \cdot 7$ .

Factor $9$ out of $63$ .

$x = \frac{- 3 \pm i \cdot \sqrt{9 (7)}}{2 \cdot - 2}$

Rewrite $9$ as $3^{2}$ .

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

Pull terms out from under the radical.

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

Move $3$ to the left of $i$ .

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

Multiply $2$ by $- 2$ .

$x = \frac{- 3 \pm 3 i \sqrt{7}}{- 4}$

Simplify $\frac{- 3 \pm 3 i \sqrt{7}}{- 4}$ .

$x = \frac{3 \pm 3 i \sqrt{7}}{4}$