Find All Complex Solutions 7z^2+8z+3=3z^2

$7 z^{2} + 8 z + 3 = 3 z^{2}$

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

Subtract $3 z^{2}$ from both sides of the equation.

$7 z^{2} + 8 z + 3 - 3 z^{2} = 0$

Subtract $3 z^{2}$ from $7 z^{2}$ .

$4 z^{2} + 8 z + 3 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = 8$ , and $c = 3$ into the quadratic formula and solve for $z$ .

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

Simplify.

Simplify the numerator.

Raise $8$ to the power of $2$ .

$z = \frac{- 8 \pm \sqrt{64 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$z = \frac{- 8 \pm \sqrt{64 - 16 \cdot 3}}{2 \cdot 4}$

Multiply $- 16$ by $3$ .

$z = \frac{- 8 \pm \sqrt{64 - 48}}{2 \cdot 4}$

Subtract $48$ from $64$ .

$z = \frac{- 8 \pm \sqrt{16}}{2 \cdot 4}$

Rewrite $16$ as $4^{2}$ .

$z = \frac{- 8 \pm \sqrt{4^{2}}}{2 \cdot 4}$

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

$z = \frac{- 8 \pm 4}{2 \cdot 4}$

Multiply $2$ by $4$ .

$z = \frac{- 8 \pm 4}{8}$

Simplify $\frac{- 8 \pm 4}{8}$ .

$z = \frac{- 2 \pm 1}{2}$

The final answer is the combination of both solutions.

$z = - \frac{1}{2}, - \frac{3}{2}$