<s>[INST] solution set of the inequality $\frac{x^2-1}{(x+2)(x+3)}>2$ 
Question: Find the solution set of the inequality $$\frac{x^2-1}{(x+2)(x+3)}>2$$

From the answer given in the previous problem I got this:
First $x\neq -2,-3$. solving the equation I get $-(2\sqrt{3}+5)<x<(2\sqrt{3}-5)$.
Is this ok?
 [/INST] $$\frac { x^{ 2 }-1 }{ \left( x+2 \right) \left( x+3 \right)  } >2\\ \frac { x^{ 2 }-1 }{ \left( x+2 \right) \left( x+3 \right)  } -2>0\\ \frac { -{ x }^{ 2 }-10x-13 }{ \left( x+2 \right) \left( x+3 \right)  } >0\\ \frac { { x }^{ 2 }+10x+13 }{ \left( x+2 \right) \left( x+3 \right)  } <0\\ \frac { \left( x+2\sqrt { 3 } +5 \right) \left( x+5-2\sqrt { 3 }  \right) \left( x+2 \right) \left( x+3 \right)  }{ { \left( x+2 \right)  }^{ 2 }{ \left( x+3 \right)  }^{ 2 } } <0\\ \left( x+2\sqrt { 3 } +5 \right) \left( x+5-2\sqrt { 3 }  \right) \left( x+2 \right) \left( x+3 \right) <0\\ \\ \\ \\ $$can you take from here? (your answer is wrong by the way)
</s>