jmc

We can start by factoring the polynomials in the numerator and denominator, which gives us $$\frac{x(2x-3)}{x(x-1)}+5x-11=\frac{(x+1)(3x+2)}{(x+1)(x-1)}.$$If $x\ne 0$ and $x\ne -1$, we can cancel out some factors to get $$\frac{2x-3}{x-1}+5x-11=\frac{3x+2}{x-1}.$$Moving the fractional terms to one side gives us $$5x-11=\frac{x+5}{x-1}.$$Now we can eliminate the denominator by multiplying by $x-1$ on both sides (as long as $x\ne 1$) and then move all the terms to one side, $$5x^2-17x+6=0.$$Factoring gives us $$(x-3)(5x-2)=0.$$Hence, $x$ must be $\boxed{3}$ or $\boxed{\frac{2}{5}}$.