# How do you find the zeros, real and imaginary, of  y=-6x^2-2x+2  using the quadratic formula?

Jul 2, 2016

The polynomial is of order 2 so has two roots. These are real, distinct and given by:

$x = - \frac{1}{6} - \frac{\sqrt{13}}{6}$ or $x = - \frac{1}{6} + \frac{\sqrt{13}}{6}$

#### Explanation:

The quadratic formula, for an equation of the form $a {x}^{2} + b x + c = 0$ is given by:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

For this question a = -6, b = -2, c = 2.

$x = \frac{2 \pm \sqrt{4 - 4 \left(- 6\right) \left(2\right)}}{2 \left(- 6\right)}$

$x = \frac{2 \pm \sqrt{52}}{- 12} = \frac{2 \pm 2 \sqrt{13}}{- 12}$

$x = - \frac{1}{6} - \frac{\sqrt{13}}{6}$ or $x = - \frac{1}{6} + \frac{\sqrt{13}}{6}$