Answer
{$-\frac{4}{3},-1$}
Work Step by Step
First, we add the fractions on the left hand side by taking their LCM. Upon inspection, the LCM is found to be $x(x+2)$:
$\frac{1}{x+2}-\frac{2}{x}=3$
$\frac{1(x)-2(x+2)}{x(x+2)}=3$
$\frac{x-2x-4}{x(x+2)}=3$
$\frac{-x-4}{(x^{2}+2x)}=3$
Now, we cross multiply the two fractions in order to create a linear equation:
$\frac{-x-4}{(x^{2}+2x)}=\frac{3}{1}$
$1(-x-4)=3(x^{2}+2x)$
$-x-4=3x^{2}+6x$
$3x^{2}+6x=-x-4$
$3x^{2}+6x+x+4=0$
$3x^{2}+7x+4=0$
Now, we use rules of factoring trinomials to solve the equation:
$3x^{2}+7x+4=0$
$3x^{2}+3x+4x+4=0$
$3x(x+1)+4(x+1)=0$
$(x+1)(3x+4)=0$
$(x+1)=0$ or $(3x+4)=0$
$x=-1$ or $x=-\frac{4}{3}$
Therefore, the solution is {$-\frac{4}{3},-1$}.
