The absolute values flip at $x=-4$ and $x=-2$, giving three regions.

Case I — $x<-4$: both $x+4$ and $x+2$ negative. Equation becomes $-x(x+4)-3(x+2)+10=0\Rightarrow x^2+7x-4=0$. Roots $x=\frac{-7\pm \sqrt{65}}{2}$. Only the more negative root $\frac{-7-\sqrt{65}}{2}\approx -7.53$ lies in $x<-4$. One solution.

Case II — $-4\le x<-2$: $x+4\ge 0,x+2<0$. Equation becomes $x(x+4)-3(x+2)+10=0\Rightarrow x^2+x+4=0$. Discriminant $=1-16<0$. No real solution.

Case III — $x\ge -2$: both non-negative. Equation becomes $x(x+4)+3(x+2)+10=0\Rightarrow x^2+7x+16=0$. Discriminant $=49-64<0$. No real solution.

Total: 1 distinct real solution.