# How do you find the roots, real and imaginary, of y= -3x^2 -+5x-2 using the quadratic formula?

Feb 4, 2018

${x}_{1} = \frac{6}{- 6} = - 1$

${x}_{2} = \frac{4}{- 6} = - \frac{2}{3}$

#### Explanation:

The quadratic formula states that if you have a quadratic in the form $a {x}^{2} + b x + c = 0$, the solutions are:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case, $a = - 3$, $b = - 5$ and $c = - 2$. We can plug this into the quadratic formula to get:
$x = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - 4 \cdot - 3 \cdot - 2}}{2 \cdot - 3}$

$x = \frac{5 \pm \sqrt{25 - 24}}{- 6} = \frac{5 \pm 1}{- 6}$

${x}_{1} = \frac{6}{- 6} = - 1$

${x}_{2} = \frac{4}{- 6} = - \frac{2}{3}$