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

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Answer 1 · 2018-06-26T05:46:16

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#