Question 919169
NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by 
h(t)=−4.9t2+160t+290.
 Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
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Solve:: -4.9t^2 + 160t + 290 = 0
t = [-160 +- sqrt(160^2 - 4*-4.9*290)]/(2*-4.9)
Positive Solution::
t = 34.37 seconds
 The rocket splashes down after 34.37 seconds.
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How high above sea-level does the rocket get at its peak?
Max occurs when t = -b/(2a) = -160/(2*-4.9) = 16.33
f(16.33) = -4.9(16.33)^2 + 160(16.33) + 290 = 1696.1 meters 
The rocket peaks at 1696.1 meters above sea-level.
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Cheers,
Stan H.
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