# How do you find the zeros, real and imaginary, of y=-6x^2+3x-9 using the quadratic formula?

Jun 9, 2018

color(red)(x = 1/4 - i sqrt207/12, 1/4 + i sqrt207/12

#### Explanation:

$y = - 6 {x}^{2} + 3 x - 9$

$a = - 6 , b = 3 , c = - 9$

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

$x = \frac{- 3 \pm \sqrt{9 - 216}}{-} 12$

color(red)(x = 1/4 - i sqrt207/12, 1/4 + i sqrt207/12