# How do you find the zeros, real and imaginary, of y= -2x^2-x-19  using the quadratic formula?

May 29, 2016

Zeros of $f \left(x\right)$ are $\left\{- \frac{1}{4} + i \frac{\sqrt{151}}{4} , - \frac{1}{4} - i \frac{\sqrt{151}}{4}\right\}$

#### Explanation:

To find zeros of $f \left(x\right) = - 2 {x}^{2} - x - 19$, we need roots of the equation $- 2 {x}^{2} - x - 19 = 0$, which can be obtained using quadratic formula.

As the roots of $a {x}^{2} + b x + c = 0$ are $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Hence, roots of zeros of $- 2 {x}^{2} - x - 19 = 0$ are

$x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(- 2\right) \left(- 19\right)}}{2 \left(- 2\right)}$

= $\frac{1 \pm \sqrt{1 - 152}}{- 4} = - \frac{1}{4} \pm i \frac{\sqrt{151}}{4}$

Zeros of $f \left(x\right)$ are $\left\{- \frac{1}{4} + i \frac{\sqrt{151}}{4} , - \frac{1}{4} - i \frac{\sqrt{151}}{4}\right\}$