How to solve this $3<x^2-4<x+1$

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