How many real zeros does y = (x-8)^3 + 12 have?

To get the zeroes, we will equate the given equation with zero and solve for x as follows:
(x-8)^3 +12 = 0
(x-8)^3 = -12
x-8 = ∛-12
x - 8 ≈ -2.2894
x ≈ -2.2894 + 8 
x ≈ 5.71

Based on the above, this equation has only one real zero