How to solve this $3Solve for, x
$$3<x^2-4<x+1 $$
Attempt:
$$7<x^2<x+5 $$
This is difficult.

Solving this equation $$3x-3<6$$ it is easy.
$$3x<9$$
$$x<3$$
 A: Hint $\:$Take the inequalities separately 
$$7<x^2\iff \sqrt {7}<|x|\iff x\in (-\infty, -\sqrt 7)\cup(\sqrt7, \infty) $$
$$x^2<x+5\iff x^2-x-5<0$$ This is like considering the parabola $p:\; y=x^2-x-5$ (open to the top) and calculating its zeros which are the interval where the inequality is valid. Therefore $$x\in\bigg(\frac{1-\sqrt{21}}{2},\frac{1+\sqrt{21}}{2}\bigg)$$ Can you finish now? I got 

 $$x\in\bigg(\sqrt7,\frac{1+\sqrt{21}}{2}\bigg)$$

A: You can see that $3 < x^2 - 4 < x+1$ is equal to $x | 3< x^2 - 4 \cap  x | x^2 -4 < x+1 $ and solve for each set
$3 < x^2-4 \leftrightarrow 7 < x^2 \leftrightarrow \sqrt{7} < | x | \leftrightarrow x \in ( - \infty , - \sqrt{7}) \cup ( \sqrt{7} , \infty )$
$ x^2 - 4 < x+1 \leftrightarrow x^2 - 4-(x+1)<0 \leftrightarrow x^2 - x - 5 < 0 \leftrightarrow (x-\frac{1- \sqrt{21}}{ 2 })(x-\frac{1+ \sqrt{21} }{ 2 }) < 0 \leftrightarrow x \in (\frac{1- \sqrt{21}}{ 2 }, \frac{1+ \sqrt{21} }{ 2 })$
You can intersect both solution sets and get the solution of your problem.
