# How do you solve the quadratic using the quadratic formula given 3a^2-4a-4=0 over the set of complex numbers?

Aug 15, 2016

This equation has 2 real roots: ${a}_{1} = - \frac{2}{3}$ and ${a}_{2} = 2$

#### Explanation:

To solve this equation using the quadratic formula first we calculate the determinant:

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

$\Delta = {\left(- 4\right)}^{2} - 4 \cdot 3 \cdot \left(- 4\right) = 16 + 48 = 64$

$\Delta > 0$ so the equation has 2 real roots:

${x}_{1} = \frac{- b - \sqrt{\Delta}}{2 a}$ and ${x}_{2} = \frac{- b + \sqrt{\Delta}}{2 a}$

${a}_{1} = \frac{4 - 8}{6} = - \frac{2}{3}$

${a}_{2} = \frac{4 + 8}{6} = 2$

Now we can write the answer:
The equation has 2 solutions ${a}_{1} = - \frac{2}{3}$ and ${a}_{2} = 2$.