Algebra 2 math help

Question

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of a second.

y=-16x^2+170x+61

Answer 1

-16t^2 +170x +61 = 0

x = -170/-32 + (1/32)sqr(170^2 +64(61))

= (170 + 181.119)/32

= about 10.972 seconds to hit the ground

rough check on the answer, let x=11

-16(11)^2 +170(11) +61

= -1936 + 1870+61

=-1936 +1931 = about zero