6.141. $\frac{24}{x^{2}+2 x-8}-\frac{15}{x^{2}+2 x-3}=2$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x^{2}+2 x-8 \neq 0, \\ x^{2}+2 x-3 \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \neq-4, \\ x \neq 2, \\ x \neq-3, \\ x \neq 1 .\end{array}\right.\right.$

Let $x^{2}+2 x=y$.

We have $\quad \frac{24}{y-8}-\frac{15}{y-3}-2=0 \Leftrightarrow \frac{y(2 y-31)}{(y-8)(y-3)}=0, \quad$ from which $\quad y_{1}=0$, $y_{2}=\frac{31}{2}$. Then $x^{2}+2 x=0, x_{1}=0, x_{2}=-2$, or $x^{2}+2 x=\frac{31}{2}$, from which $x_{3,4}=\frac{-2 \pm \sqrt{66}}{2}$.

Answer: $x_{1}=0, x_{2}=-2, x_{3,4}=\frac{-2 \pm \sqrt{66}}{2}$.