Question 595766
An arrow shot into the air is 160t-4.9t^2 meters above the ground t seconds after its released. During what period(s) of time is the arrow below 117.6 meters? Round your answer to the nearest .01 second. 
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f(t)=160t-4.9t^2
117.6=160t-4.9t^2
4.9t^2-160t+117.6=0
use following quadratic formula to solve for t
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
a=4.9, b=160, c=117.6
t=.75 and 31.90 seconds
answer A