Question

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

Answer 1

The quadratic formula for a function of the form

`y=ax^2+bx+c`

is

`x=(-b+-sqrt(b^2-4ac))/(2a)`

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

Substitute.

`x=(-(-1)+-sqrt((-1)^2-4(1)(81)))/(2(1))`

Simplify

`x=(1+-sqrt(-323))/2 rarr (1+-sqrt(323i^2))/2 rarr (1+-isqrt(323))/2`