I cant seem to find a way to actually find any solutions to this....any suggestions?
$\displaystyle 3/(x^2+x)-1/(x+5)=0$
NO.
$\displaystyle \frac{3(x+5)-x(x+1)}{(x^2+x)(x+5)} $ is what you had gotten.
according to your question, this term = 0, so you have,
$\displaystyle \frac{3(x+5)-x(x+1)}{(x^2+x)(x+5)} = 0$
Look at what Wilmer has said in the post above. That gives you:
$\displaystyle 3(x+5)-x(x+1) = 0$
$\displaystyle 3x+15-x^2-x=0$
$\displaystyle 2x+15-x^2=0$
$\displaystyle x^2-2x-15=0$
Solve this quadratic equation for the value of x!