A model rocket is launched straight upward from the side of a 212-ft cliff. The initial velocity is 86 ft/sec. The height of the rocket h(t) is given by: h(t)=-16t^2+86t+212

where h(t) is measured in feet and t is the time in seconds. Use the quadratic formula to find the times (t) at which the rocket is 320 ft. above the ground. Round your answers to the nearest hundredth of a second.

1 answer

just solve for t in

-16t^2+86t+212 = 320