jmc

We first notice that we can simplify the fraction: $$\frac{x^3+2x^2}{x^2+3x+2}=\frac{x^2(x+2)}{(x+1)(x+2)}=\frac{x^2}{x+1},$$provided that $x\ne -2.$ Therefore, we have $$\frac{x^2}{x+1}+x=-6.$$Multiplying both sides by $x+1$ gives $$x^2+x(x+1)=-6(x+1),$$or $$2x^2+7x+6=0.$$This equation factors as $$(2x+3)(x+2)=0,$$so $x=-\frac{3}{2}$ or $x=-2.$ But, as we said before, $x=-2$ is impossible because it makes the denominator of the fraction equal to zero. Therefore, the only valid solution is $x=\boxed{-\frac{3}{2}}.$