First of all, if the position functions are really as you typed them:
s = 832 - 2.6t^2
and
s = 832 - 16t^2

then your derivatives are wrong because there is no t in the first term of each equation.

On the other hand, the problem, as you posted it, doesn't say that s is the MAXIMUM height, and it also isn't clear whether there are two different s's, or only one value of s...

On the other hand, there really should be a t in the first term, since the usual position function is s = v_{0}t - .5at^2
So should it say s = 832t - 16t^2???

On the other hand, maybe the trick is that the t is "hidden" such that 832 = v_{0}t and both v_{0} and t are unknown...

On the other hand, that acceleration for the moon doesn't really make sense either; free fall acceleration near the surface of the moon is approx. 1.6 m/s^2...

You know, you can make your work really easier (no need for calculus !).
s = 832t - 2.6t^2
s is displacement.
now, displacement will be equal to zero in only two cases (in projectiles) :
1-the object didn't move yet
2-the object went up and back down to the earth's surface.
So all you have to do is to substitue the value of s with 0, and solve the equation .

And BTW, if i understood ur first way of solving this right, then i think there is something wrong about it. You see the velocity of the bullet will not be 0 when it reaches the surface, it will actually be equal in magnitude (and opposite in direction) of the initial velocity.