A model rocket is launched upward with an initial velocity of 200 feet per second. The height, in feet, of the rocket t seconds after the launch is given by h = −16t2 + 200t. How many seconds after the launch will the rocket be 380 feet above the ground? Round to the nearest tenth of a second. (Enter your answers as a comma-separated list.)

1 answer

just solve

-16t^2 + 200t = 380
-16t^2 + 200t - 380 = 0
Now just use the quadratic formula to find the roots.