Algebra 2
posted by Marcia .
A bug is placed on the road. t seconds after it is put down, the function is p(t) = 16tsquared + 64t, where p(t) is measure in feet. At what moment will the bug be 14 feet from where it began?

p(t) same as f(x) basically just put in 14 for 7 in the equation but don't put it in for the t in p(t) solve out and that is your answer
p(t)=16(14)^2 + 64(14) 
oh whoops i read the question wrong scratcch that

oh okay here it is sorry about the above;
p(t)=  16t^2 +64t
14=  16t^2 + 64t bring 14 to other side and factor or use the equation b = or minus square root of (b  4AC)and divide by 2A and you should get what t is 
14=  16t^2 + 64t
16 t^2  64 T + 14 = 0
8 t^2 32 t + 7 = 0
t = [32 +/sqrt (1024224)]/16
= [32 +/ sqrt(800)]/16
= [32 +/ 20sqrt 2]/16
= [ 32 +/ 28.3]/16
= 3.76 on the way down
and
= .231 on the way up
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