Answer :
[tex]x^2\leq16\\x\leq 4 \wedge x\geq -4\\x\in\langle-4,4\rangle[/tex]
different approach:
[tex]x^2\leq16\\x^2-16\leq0\\(x-4)(x+4)\leq0[/tex]
see attachment
[tex]x\in\langle-4,4\rangle[/tex]

[tex]~~~~~~x^2 \leq 16\\\\\implies -\sqrt{16} \leq x\leq \sqrt{16}\\\\\implies -4\leq x \leq 4\\\\\text{Interval notation:}~ [-4,4][/tex]