Помогите решить неравенство, плз. (x - 3)(x^2 + 4)^(1/2) меньше = x^2 - 9

[latex](x-3)\sqrt{x^2+4}\leq (x-3)(x+3) \\; (x-3)\sqrt{x^2+4}-(x-3)(x+3) \leq 0 \\ (x-3)(\sqrt{x^2+4}-x-3}\leq 0 \\ x = 3 ; \sqrt{x^2+4}=x+3 \\ x+3\ \textgreater \ 0; x\ \textgreater \ -3 \\ x^2+4-(x+3)^2=0 \\ x=-\frac{5}{6} \\ -+- \\ x \in (-\infty;-\frac{5}{6}] \cup [3;+\infty)[/latex]