I'm 12 so please don't laugh at my math. 

Let's assume that an object has KE expressed by velocity $v$ and mass $m$. 
If work $w$ has been done to this object such that it's new speed (same direction) is $s$. There are 2 ways(I know) to calculate $w$. The first is 
$ \frac {1}{2} ms^2 -  \frac {1}{2} mv^2$
Now, I know what you are thinking, what other way do you need?
Anyway, the other way is to(assuming you know distance and acceleration)
$w=mda $
then
$d= \frac {1}{2}at^2 + vt  $ 
next

$\frac{2d}{a}= t^2 + \frac {2v}{a}t$
next
$\frac {2d}{a} +\frac {v^2}{a^2} = t^2 + \frac {2v}{a}t + \frac {v^2}{a^2}$ 
next

$\frac {2d}{a} +\frac {v^2}{a^2} = (t+\frac {v}{a})^2$
next

$\sqrt {\frac {2d}{a} +\frac {v^2}{a^2}} = t + \frac{v}{a}$ 
next

$\sqrt {\frac {2da+v^2}{a^2} = t + \frac{v}{a}$
next
$\frac {sqrt {2da+v^2}}{a} = t + \frac{v}{a}$ 
next

$\frac {sqrt {2da+v^2}-v}{a} = t $ 
next

and  the speed is a*t which is equal to 
${sqrt {2da+v^2}-v}$
remember that 

$ da = \frac {w}{m}$
so 

$ s = ${sqrt {\frac {2w}{m}+v^2}-v}$

so 
$ (s+v)^2 = \frac{2w}{m} +v^2$

so 
$ \frac{1}{2}m((s+v)^2-v^2) = w $
simplify and it will be
$ w= \frac {1}{2} ms^2 + \frac{1}{2} mvs
which isn't quite the other formula. Why is it that my formula doesn't work?