How many seconds after the launch will the rocket be 300 feet above the ground? Round to the nearest tenth of a second

Question

A model rocket is launched upward with an initial velocity of 220 feet per second. The height, in feet, of the rocket t seconds after the launch is given by h = −16t² + 220t.

 How many seconds after the launch will the rocket be 300 feet above the ground? Round to the nearest tenth of a second.

Answer 1

300 = -16t^2 +220t

16t^2 -220t +300=0

8t^2 -110t + 150=0

4t^2 - 55t + 75 =0

t = 55/8 + or - (1/8)sqr(55^2 -16(75)) =6.875 + or - 5.340 = 1.53 or 12.22 seconds