Quick yes or no question

Okay so the problem states:

If an object is thrown upward with an initial velocity of 12 ft per sec, then it's height (in feet) is given by h(t)=-12t^2+72t after time t seconds. What is it's maximum height?

My question is do I just plug in 12 ft for t or h?

No, notice the equation is for an inverted parabola since a = -12 < 0,the vertex is the global max so:
t = -72/-24 = 3 => time to get to max h
h(3) = 108 ft => max h

Edit: using standard quadratic equation:
y = ax^2 + bx + c => the coordinates of the vertex are: (-b/2a , y(-b/2a)):
h(t) = -12t^2 + 72t
t = -b/2a = -72/-24 = 3 sec
h(3) = -12*(3)^2 + 72 * 3
= -108 + 216 = 108 ft

If you have already studied the derivatives, you can take the derivative of the function and set it to zero, as:
h(t) = -12t^2 + 72t
dh/dt = -24t + 72 => set to zero solve for t:
-24t + 72 = 0
-24(t - 3) = 0
t = 3 => time when the object reaches max h, plug in and find max h.(already done above)

If you have any questions, please let me know.

no

there are several ways to solve it, but yours isn't one of them