How do I solve (e^-0.2x)-(e^-0.75 x=0.25?

It is also possible that the equation in question is meant to be:

$$e^{-0.2x}-e^{-0.75x}=0.25$$ 

In this case there are two solutions as can be seen from the graph below:

These solutions can be obtained numerically.  A simple iterative approach is to write the equation in the following form:

$$x_{n+1}=\frac{-1}{0.2}ln(0.25+e^{-0.75x_n})$$ 

Then let x0 = 1, say, and use the formula above to iterate several times.  The result is:

x ≈ 6.812073

By rewriting the iteration formula as:

$$x_{n+1}=\frac{-1}{0.75}ln(e^{-0.2x_n}-0.25)$$

we can iterate to the other solution:

x ≈ 0.6023554654

How about with the site calculator?

$${{\mathtt{e}}}^{\left(-{\mathtt{0.2}}\right)}{\mathtt{\,\times\,}}{\mathtt{0.25}}{\mathtt{\,-\,}}{{\mathtt{e}}}^{\left(-{\mathtt{0.75}}\right)}{\mathtt{\,\times\,}}{\mathtt{0.25}} = {\mathtt{0.086\: \!591\: \!050\: \!084\: \!241\: \!8}}$$

I am not sure if this is actually what you meant - you must be careful to check the brackets are in the right place.

(It is what you asked i believe)

Maybe you meant this:

e^(-.2) x  - e^(-.75) x  = .25 ??

I'm going to assume so......So we have

0.8187307530779818 x - 0.4723665527410147 x = .25

0.3463642003369671 x = .25        So x = .25 / 0.3463642003369671 = 0.7217836016446927047

Chris thinks you meant

$$(e^{-0.2}\times x) - (e^{-0.75}\times x)=0.25$$

Yes, this probably is what you meant.  Your title is missing a bracket so there is plenty of room for confusion!

I read in an extra x (without even realising) and Chris read in an extra bracket.  I now see your main text does have the bracket!