h=13t²-5t. This can be written h=13(t²-5t/13)⇒h=13(t²-5t/13+(5/26)²-(5/26)²).
h=13(t-5/26)²-25/52. In this form it’s possible to see that when t=5/26=0.19 sec, approximately, the maximum depth is 25/52=0.48m approx, because all other values of t make h less negative, i.e., the diver is closer to the surface. And when h=t(13t-5)=0 (on the surface) t=0 (initial condition on commencement of the dive) and t=5/13=0.38 seconds, which is twice the time it took to reach maximum depth.