0\\ (x-6)^2>0\\ x\in\mathbb{R}\setminus\{6\}\\\\ \log_{\log _{3}2}(x^2-12x+36)\geq 0\\ x^2-12x+36\leq (\log_3 2)^0\\ x^2-12x+36\leq1\\ x^2-12x+35\leq0\\ x^2-7x-5x+35\leq0\\ x(x-7)-5(x-7)\leq0\\ (x-5)(x-7)\leq0\\ x\in\langle5,7\rangle\\\\ x \in\langle5,7\rangle\wedge x\in\mathbb{R}\setminus\{6\}\\x\in\langle5,6)\cup(6,7\rangle\\\\ x\in(\langle5,6)\cup(6,7\rangle) \cap \mathbb{Z}\\ x=\{5,7\}\\\\ 5+7=\underline{12} " alt="\\x^2-12+36>0\\ (x-6)^2>0\\ x\in\mathbb{R}\setminus\{6\}\\\\ \log_{\log _{3}2}(x^2-12x+36)\geq 0\\ x^2-12x+36\leq (\log_3 2)^0\\ x^2-12x+36\leq1\\ x^2-12x+35\leq0\\ x^2-7x-5x+35\leq0\\ x(x-7)-5(x-7)\leq0\\ (x-5)(x-7)\leq0\\ x\in\langle5,7\rangle\\\\ x \in\langle5,7\rangle\wedge x\in\mathbb{R}\setminus\{6\}\\x\in\langle5,6)\cup(6,7\rangle\\\\ x\in(\langle5,6)\cup(6,7\rangle) \cap \mathbb{Z}\\ x=\{5,7\}\\\\ 5+7=\underline{12} " align="absmiddle" class="latex-formula">