Question

x(t) = t^3 - 12t^2 + 36t
Velocity = x'(t) = 3y^3 - 24t + 36 = 0
x'(t) = t^2 - 8t + 12 = 0
x'(t) = t^2 - 6t - 2t + 12 = 0
x'(t) = t(t - 6) - 2(t - 6) = 0
x'(t) = (t - 2)(t - 6) = 0
t = 2 and t = 6 must be the roots.

Is my answer of D "2 and 6" correct based on this work?

Answer 1

Yes you have the right idea. You find the derivative to get the velocity function and then determine when the velocity is zero. That turns out to be at t = 2 and t = 6. Both are correct. The final answer is correct.

The only flaw I see is that you wrote 
x'(t) = 3y^3 - 24t + 36
on line 2 when it should be
x'(t) = 3t^2 - 24t + 36

Other than that it looks perfect.