Find the Roots (Zeros) -8x+30=-3x^2+5x

$- 8 x + 30 = - 3 x^{2} + 5 x$

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- 3 x^{2} + 5 x = - 8 x + 30$

Step 2

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

Add $8 x$ to both sides of the equation.

$- 3 x^{2} + 5 x + 8 x = 30$

Add $5 x$ and $8 x$ .

$- 3 x^{2} + 13 x = 30$

Step 3

Subtract $30$ from both sides of the equation.

$- 3 x^{2} + 13 x - 30 = 0$

Step 4

Use the quadratic formula to find the solutions.

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

Step 5

Substitute the values $a = - 3$ , $b = 13$ , and $c = - 30$ into the quadratic formula and solve for $x$ .

$\frac{- 13 \pm \sqrt{13^{2} - 4 \cdot (- 3 \cdot - 30)}}{2 \cdot - 3}$

Step 6

Simplify.

Simplify the numerator.

Raise $13$ to the power of $2$ .

$x = \frac{- 13 \pm \sqrt{169 - 4 \cdot - 3 \cdot - 30}}{2 \cdot - 3}$

Multiply $- 4 \cdot - 3 \cdot - 30$ .

Multiply $- 4$ by $- 3$ .

$x = \frac{- 13 \pm \sqrt{169 + 12 \cdot - 30}}{2 \cdot - 3}$

Multiply $12$ by $- 30$ .

$x = \frac{- 13 \pm \sqrt{169 - 360}}{2 \cdot - 3}$

Subtract $360$ from $169$ .

$x = \frac{- 13 \pm \sqrt{- 191}}{2 \cdot - 3}$

Rewrite $- 191$ as $- 1 (191)$ .

$x = \frac{- 13 \pm \sqrt{- 1 \cdot 191}}{2 \cdot - 3}$

Rewrite $\sqrt{- 1 (191)}$ as $\sqrt{- 1} \cdot \sqrt{191}$ .

$x = \frac{- 13 \pm \sqrt{- 1} \cdot \sqrt{191}}{2 \cdot - 3}$

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

$x = \frac{- 13 \pm i \sqrt{191}}{2 \cdot - 3}$

Multiply $2$ by $- 3$ .

$x = \frac{- 13 \pm i \sqrt{191}}{- 6}$

Simplify $\frac{- 13 \pm i \sqrt{191}}{- 6}$ .

$x = \frac{13 \pm i \sqrt{191}}{6}$

Step 7