Question

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

Answer 1

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

Explanation:

#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)#