What are the zeros of f (x) = x^2-12+36

x = 6

Step-by-step explanation:

I think it's supposed to be

y = x² - 12x + 36

Set the equation equal to zero and solve ...

0 = x² - 12x + 36

Now factor. What 2 numbers add to - 12, and multiply to 36?

-6 and - 6, so our factors are

0 = (x - 6) (x - 6)

So

0 = x - 6, then

6 = x