The set of all real number x such that $||3-x|-| x+2||=5$

The set of all real number x such that $$||3-x|-| x+2||=5$$ is??

My approach:- $(|| 3-x|-| x+||2)^{2}=25$

$\Leftrightarrow(3-x)^{2}+(x+2)^{2}-2|3-x||x+2|=25$

$\Rightarrow x^{2}-x-\left|-x^{2}+x+6\right|=6$

What to do next?.....

Edit after seeing the comments I got the idea,

So, it is clear that $-x^{2}+x+6<0,$ i.e. $-x^{2}+x+6 \geq 0$

$(x-3)(x+2) \geq 0$

So, $x \leq-2 \& x \geq 3$

$\therefore \mathrm{x} \in(-\infty,-2] \cup[3, \infty)$

Correct me ,If I am wrong

There is no need for squaring. Squaring has a risk of introducing extraneous solutions. Instead we can say \begin{align*} |3-x|-|x+2|&=\pm 5 \tag{1} \end{align*} Now consider three cases:

1. $x \geq 3$. In this case equation (1) becomes $$-3+x-x-2=\pm 5 \implies -5=5 \text{ OR } \color{red}{-5=-5}$$ Because of the latter, $\color{red}{x \geq 3}$ is a solution interval.

Now do the same for $-2 \leq x < 3$ and $ x < -2$.