Is there a simpler way to solve for time in this algebraic equation?

[jKotha]

hey hey.. I need help again. I can't believe I've forgotten my simple algebra. Anyway.. is there a way to solve the following without using the quadratic equation?

0 = (22)t + 1/2(-9.8)t^2

I'm tryin to isolate t, so i can solve for time. I know you can factor it out as follows..

0 = [22 + 1/2(-9.8)t]t

Thats where I am stuck... how do i get this in the form.. t = xxxxx

Thanks for the help..

 

[courtrigrad]

22t -4.9t^{2} = 0 \rightarrow t(22-4.9t) = 0

t = 0 or 22-4.9t = 0, t = 4.49

 

[jKotha]

where did the 4.9 come from? nevermind.. figured that out.. THANKS!>. i feel really dumb now because i should have known that.. sometimes i complicate myself too much.

 

[QuantumCrash]

Thats usually the best method but for this one it is not necessary.
Picking up from u
0 = [22 + 1/2(-9.8)t]t
Therefore
t = 0 and 22 + 1/2 (-9.8)t =0
finally:
t= 0, 4.49s

I think it is the second answer that you probably want. Since this type of question is probably asking when the onject will return to its original position.