jmc

We notice that the denominator on the left factors, giving us $$\frac{2x+4}{(x-1)(x+5)}=\frac{2-x}{x-1}.$$As long as $x\ne 1$ we are allowed to cancel $x-1$ from the denominators, giving $$\frac{2x+4}{x+5}=2-x.$$Now we can cross-multiply to find $$2x+4=(2-x)(x+5)=-x^2-3x+10.$$We simplify this to $$x^2+5x-6=0$$and then factor to $$(x-1)(x+6)=0.$$Notice that since $x-1$ is in the denominator of the original equation, $x=1$ is an extraneous solution. However $x=\boxed{-6}$ does solve the original equation.