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

May 18, 2016

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

$- \frac{1}{4} + i \frac{\sqrt{23}}{4}$ and $- \frac{1}{4} - i \frac{\sqrt{23}}{4}$

#### Explanation:

If $y = a {x}^{2} + b x + c$, quadratic formula states that zeros are given by

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

Hence, in $y = - 2 {x}^{2} + x - 3$, zeros are

$\frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot \left(- 2\right) \cdot \left(- 3\right)}}{2 \cdot 2}$

= $\frac{- 1 \pm \sqrt{1 - 24}}{2 \cdot 2}$

or $\frac{- 1 \pm \sqrt{- 23}}{4} = - \frac{1}{4} \pm i \frac{\sqrt{23}}{4}$