math
posted by sabrina on .
A model rocket is launched with an initial velocity of 250 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 250t.
How many seconds after launch will the rocket be 730 ft above the ground? Round to the nearest hundredth of a second.

730=16t^2+250t
16t^2250t+730=0
Using the quadratic equation...
t= (250+sqrt (250^24*16*730))/32
solve for t