What are all the real zeroes of y = (x - 12) ^3 - 7

You might be right.

Let y = 0

(x - 12) ^3 - 7 = 0 Subtract 7 from both sides

(x - 12) ^3 = 7 Take the cube root of both sides.

x - 12 = cuberoot (7) Add 12 to both sides

x = 12 + cuberoot (7)

Are the other two real or imaginary? The quickest way to find out the answer is to graph the original equation. It has the shape of something that crosses the x axis but once. So the other two roots are imaginary.

My calculator says that the real root is 13.91

The two complex ones are 11.04 + / - 1.66i which of course is not real.

y = (x - 12) ^3 - 7

set y=0

0 = (x - 12) ^3 - 7

add 7 to each side

7 = (x - 12) ^3

take the cube root of each side

7^ 1/3 = x-12

add 12 to each side

12 + 7^ (1/3) = x

this is the only real root

the other 2 are imaginary