# How do you solve the quadratic using the quadratic formula given 3t^2+8t+5=-2t^2 over the set of complex numbers?

Dec 24, 2016

The solutions are $S = \left\{- \frac{4}{5} + \frac{3}{5} i , - \frac{4}{5} - \frac{3}{5} i\right\}$

#### Explanation:

We need

$\Delta = {b}^{2} - 4 a c$

and $x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

Let's rewrite the equation

$3 {t}^{2} + 8 t + 5 = - 2 {t}^{2}$

$5 {t}^{2} + 8 t + 5 = 0$

$\Delta = {8}^{2} - 4 \cdot 5 \cdot 5 = 64 - 100 = - 36$

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

$t = \frac{- 8 \pm 6 i}{10}$

$t = - \frac{4}{5} \pm \frac{3}{5} i$