How can the time variable be isolated in the equation D=VoT+1/2AT^2?

[rigory]

Homework Statement

For quite a while I have been working on this equation and attempting to isolate time, however i have so far been unable to do so.

Homework Equations

D=volt+1/2AT^2

The Attempt at a Solution

I've been going through and keep coming up with T on both sides of the equation. I'd very much appreciate any help in this area.

 

[fluidistic]

It seems to me a quadratic function. If it is so, check out http://en.wikipedia.org/wiki/Quadratic_function, the "Roots" part.

 

[Gregg]

Homework Statement

For quite a while I have been working on this equation and attempting to isolate time, however i have so far been unable to do so.

Homework Equations

D=volt+1/2AT^2

The Attempt at a Solution

I've been going through and keep coming up with T on both sides of the equation. I'd very much appreciate any help in this area.

to isolate time in $s=vt+1/2at^2$

$$1/2AT^2 + volt - D = 0$$a=1/2A
b=Vo
c=-D

$$X = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$$

we get

$$T = \frac{-Vo + \sqrt{(Vo)^2 - 4(1/2A)(-D)}}{2(1/2A)}$$

 

[rigory]

thanks, i think that will work, I've been trying to program it into a graphing calculator, but couldn't find a way to get it.

 

[fluidistic]

Well you're lucky enough Gregg did all the job for you. (By the way it's forbidden according to the forum's rules).
You really should learn how to find roots of any quadratic function. This is a not so hard thing to accomplish and very useful. (Don't use calculator if possible.)