I'm truly stumped on how to solve for "x" in the following equation: 

(2/5-x) + 4 = 6/x-5

I've tried solving this in various different ways, but the most
logical seems to be to take care of the left side first, by
multiplying both operands on the left by (5-x), giving:

(2 + 4(5-x))/5-x = 6/x-5

From this I wind up with (22 - 4x)/5-x = 6/x-5

Now it would seem logical to cross-multiply, but then you wind up with
x^2 in the numerator, and I can't seem to get rid of it.

I've also tried multiplying both sides of the equation by the LCD
(5-x)(x-5), which leads to another overly complex equation. I've also
tried moving things from one side of the = to another first, then
working things out from there, which is also overly complex.

This equation was given to me on a review exam in preparation for a
final exam, and the answer given to the problem is "7", which, when I
plug it in appears to work just fine -- I just need to understand how
that answer was derived (solutions weren't given, only answers).

First off, let me make sure I have the equation correct.  The way I
think you intended it to look, from the steps you showed, is something
like:
2      +4 = 6
----        ---
5-x         x-5

with 5-x and x-5 completely in the denominators.  If this isn't
correct, then ignore the rest of the solution.

You were right to take care of the left-hand side, giving:

2+4(5-x)  = 6
--------   ---
  5-x      x-5

Cross-multiplying is also correct:

(x-5)[2+4(5-x)] = 6(5-x)

Now, the trick to realize is that 5-x = -1(x-5).  Put this in on the right side.

(x-5)[2+4(5-x)] = -6(x-5)

Now the x-5 cancels from each side and you're left with:

2+4(5-x) = -6

22-4x=-6

-4x=-28

x=7

Hope this helps!

You are on the right track.

Notice that when you cross-multiply, you must exclude x = 5 as a final
answer since the equation would be undefined.

You have the quadratic -4x^2 + 48x - 140 = 0 at one point.

From there, factorise:
-4x^2 + 48x - 140 = 0
=> -4(x^2 - 12x + 35) = 0
=> -4(x - 7)(x - 5) = 0
=> x = 7 or x = 5, but 5 is not in the domain of x as noted above.

Therefore, the solution is x = 7.

2         6
--- + 4 = ---;
5-x       x-5

   2         6
- --- + 4 = ---;
  x-5       x-5

     6     2
4 = --- + ---;
    x-5   x-5
    
    6+2
4 = ---;
    x-5

     8
4 = ---;
    x-5

4*(x-5) = 8;

4*x-4*5 = 8;

4*x-20 = 8;

4*x = 20+8

4*x = 28;

x = 28/4;

x = 7.