# How do you solve the quadratic using the quadratic formula given 7-8z^2=6z+16 over the set of complex numbers?

Nov 20, 2016

The solution is $S = \left\{\frac{- 3 + 3 i \sqrt{7}}{8} , \frac{- 3 - 3 i \sqrt{7}}{8}\right\}$

#### Explanation:

Rearranging the equation

$8 {z}^{2} + 6 z + 9 = 0$

Let's calculate the discriminant

$\Delta = {b}^{2} - 4 a c = 36 - 4 \cdot 8 \cdot 9 = 36 - 288 = - 252$

$\Delta < 0$, so solutions are in $\mathbb{C}$

Then,
$z = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 6 \pm i \sqrt{252}}{16}$

$= \frac{- 6 \pm 6 i \sqrt{7}}{16} = \frac{- 3 \pm 3 i \sqrt{7}}{8}$