[tex]\displaystyle\bf\\2lg(x)=lg(2x+8)\\\\lg(x^2)=lg(2x+8)\\\\\textbf{Logaritmii au aceeasi baza (10)}\\\\\implies~~x^2=2x+8\\\\x^2-2x-8=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\x_{12}=\frac{2\pm\sqrt{4+32}}{2}\\\\\\x_{12}=\frac{2\pm\sqrt{36}}{2}\\\\\\x_{12}=\frac{2\pm6}{2}\\\\x_{12}=1\pm3\\\\x_1=1+3\\\\\boxed{\bf~x_1=4}\\\\x_2=1-3\\\\\boxed{\bf~x_2=-2}[/tex]