Find All Complex Solutions 3z^2+12z-20=5z

$3 z^{2} + 12 z - 20 = 5 z$

Move all terms to the left side of the equation and simplify.

Subtract $5 z$ from both sides of the equation.

$3 z^{2} + 12 z - 20 - 5 z = 0$

Subtract $5 z$ from $12 z$ .

$3 z^{2} + 7 z - 20 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 3$ , $b = 7$ , and $c = - 20$ into the quadratic formula and solve for $z$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (3 \cdot - 20)}}{2 \cdot 3}$

Simplify.

Simplify the numerator.

Raise $7$ to the power of $2$ .

$z = \frac{- 7 \pm \sqrt{49 - 4 \cdot 3 \cdot - 20}}{2 \cdot 3}$

Multiply $- 4$ by $3$ .

$z = \frac{- 7 \pm \sqrt{49 - 12 \cdot - 20}}{2 \cdot 3}$

Multiply $- 12$ by $- 20$ .

$z = \frac{- 7 \pm \sqrt{49 + 240}}{2 \cdot 3}$

Add $49$ and $240$ .

$z = \frac{- 7 \pm \sqrt{289}}{2 \cdot 3}$

Rewrite $289$ as $17^{2}$ .

$z = \frac{- 7 \pm \sqrt{17^{2}}}{2 \cdot 3}$

Pull terms out from under the radical, assuming positive real numbers.

$z = \frac{- 7 \pm 17}{2 \cdot 3}$

Multiply $2$ by $3$ .

$z = \frac{- 7 \pm 17}{6}$

The final answer is the combination of both solutions.

$z = \frac{5}{3}, - 4$