How do you find the zeros, real and imaginary, of $y=x^2 -16x +81$ using the quadratic formula?

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Answer 1 · 2016-02-26T18:44:13

$y = 16 +-sqrt(-17)$

$y = x^2 -16x + 81 or, y = (1)x^2 + (-16)x + 81 or, y = [-(-16)+-sqrt{(-16)^2-4*1*81}]/2*1 or, y = 16+-sqrt(-68)/2 or, y = 16 +-sqrt(-17)$

How do you find the zeros, real and imaginary, of y=x^2 -16x +81y=x^2 -16x +81 using the quadratic formula?