A model rocket is launched with an initial velocity of 200 ft per second. The height h, in feet, of the rocket t seconds after the launch is given by

Question

h = −16t2 + 200t.
How many seconds after the launch will the rocket be 550 ft above the ground? Round to the nearest hundredth of a second. (Enter your answers as a comma-separated list.)

Answer 1

You want -16t^2 + 200t to be equal to 550, so ...

-16t^2 + 200t = 550
16t^2 - 200t + 550 = 0

Use the quadratic formula, you will get 2 positive answers.
Of course the smaller time would be on it way up, the larger time
will be the time on its return.