1708$\textbf{Here $\mathbf{(2+\sqrt{3})^{x^2-2x}.(2+\sqrt{3})+\left(\frac{1}{2+\sqrt{3}}\right)^{x^2-2x}.(2+\sqrt{3})=2.(2+\sqrt{3})}$}\\\\\\ \textbf{Here We have Use $\mathbf{(2+\sqrt{3})=\frac{1}{2-\sqrt{3}}}$}$\\\\\\ $\textbf{Now after Simlification, We get $\mathbf{(2+\sqrt{3})^{x^2-2x}+\left(\frac{1}{2+\sqrt{3}}\right)^{x^2-2x}=2}$}\\\\\\ \textbf{Now Let $\mathbf{(2+\sqrt{3})^{x^2-2x}=a}$, We Get }$\\\\\\ \mathbf{a+\frac{1}{a}=2\Leftrightarrow (a-1)^2=0\Leftrightarrow a=1}$\\\\\\ $\mathbf{a=(2+\sqrt{3})^{x^2-2x}=1=(2+\sqrt{3})^0\Leftrightarrow x^2-2x=0\Leftrightarrow \underline{\underline{x=0,2}}}$