jmc

We have $$\begin{array}{rl}(2x+10)(x+3)& <(3x+9)(x+8)\Rightarrow \\ 2(x+5)(x+3)& <3(x+3)(x+8)\Rightarrow \\ 2(x+5)(x+3)-3(x+3)(x+8)& <0\Rightarrow \\ (2x+10-(3x+24))(x+3)& <0\Rightarrow \\ (-x-14)(x+3)& <0\Rightarrow \\ (x+14)(x+3)& >0.\end{array}$$ This inequality is satisfied if and only if $(x+14)$ and $(x+3)$ are either both positive or both negative. Both factors are positive for $x>-3$ and both factors are negative for $x<-14$. When $-14<x<-3$, one factor is positive and the other negative, so their product is negative. Therefore, the range of $x$ that satisfies the inequality is $\boxed{(-\infty ,-14)\cup (-3,\infty )}$.