This is a freefall problem with an acceleration of -10m/s (a poor approximation of g :P)
a=g
v=gt+v0
h=gt^2/2+v0t+h0 where v0 is initial velocity and h0 is initial height.
You have:
h=-5t^2+10t+3
Since h0=3, the springboard is 3 meters above the water.
...
The diver hits the water when h=0.
5t^2-10t-3=0
Using the quadratic equation for simplicity:
t=(10±√160)/10
t=(10±4√10)/10 seconds, since t>0
t≈2.26 seconds (to nearest hundredth of a second)
So the diver hits the water about 2.26 seconds after (s)he leaves the springboard.
