# How do you find the zeros, real and imaginary, of y=x^2 -x+81 using the quadratic formula?

Jun 26, 2018

The quadratic formula for a function of the form

$y = a {x}^{2} + b x + c$

is

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

In your case a=1, b=-1 and c=81

Substitute.

$x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(1\right) \left(81\right)}}{2 \left(1\right)}$

Simplify

$x = \frac{1 \pm \sqrt{- 323}}{2} \rightarrow \frac{1 \pm \sqrt{323 {i}^{2}}}{2} \rightarrow \frac{1 \pm i \sqrt{323}}{2}$