# How do you find the zeros, real and imaginary, of y= -3x^2-8x-18 using the quadratic formula?

Jun 9, 2016

Zeros of $y = - 3 {x}^{2} - 8 x - 18$ are

$- \frac{4}{3} - 2 \sqrt{38} i$ and $- \frac{4}{3} + 2 \sqrt{38} i$

#### Explanation:

Zeros of $y = a {x}^{2} + b x + c$ are given by the quadratic formula as

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

Hence for $y = - 3 {x}^{2} - 8 x - 18$ these are given by

$\frac{- \left(- 8\right) \pm \sqrt{{\left(- 8\right)}^{2} - 4 \times \left(- 3\right) \times \left(- 18\right)}}{2 \cdot \left(- 3\right)} = \frac{8 \pm \sqrt{64 - 216}}{- 6}$

= $- \frac{4}{3} \pm \frac{\sqrt{- 152}}{6} = - \frac{4}{3} \pm i 2 \sqrt{38}$

Hence zeros of $y = - 3 {x}^{2} - 8 x - 18$ are

$- \frac{4}{3} - 2 \sqrt{38} i$ and $- \frac{4}{3} + 2 \sqrt{38} i$