ОДЗ:

[tex]\left \{ {{x^{2} -4x > 0} \atop {x+6 > 0}} \right. \\\left \{ {{x(x-4) > 0} \atop {x > -6}} \right. \\\left \{ {{x < 0 \: \: \: and \: \: \: x > 4} \atop {x > -6}} \right. \\\\-6 < x < 0\: \: \: and \: \: \: x > 4[/tex]

[tex]\lg_{} (x^{2} -4x)-\lg_{} (x+6)=0\\\lg_{} (\frac{x^{2}-4x }{x+6} )=\lg_{} (1)\\\frac{x^{2}-4x }{x+6}=1\\\frac{x^{2}-4x }{x+6}-1=0\\\frac{x^{2}-4x -x-6}{x+6}=0\\x^{2} -5x-6=0\\a=1\\b=-5\\c=-6\\D=b^{2} -4ac=(-5)^{2} -4\times1\times(-6)=25+24=49\\x_{1} =\frac{5-7}{2} =-\frac{2}{2} =-1\\x_{2} =\frac{5+7}{2} =\frac{12}{2} =6[/tex]

Корни подходят под ОДЗ, поэтому ответ: х=-1 и х=6

[tex]\displaystyle\bf\\\lg(x^{2} -4x)-\lg(x+6)=0\\\\ODZ:\\\\\left \{ {{x^{2} -4x > 0 \atop {x+6 > 0}} \right. \ \ \Rightarrow \ \ \left \{ {{x(x-4) > 0} \atop {x > -6}} \right. \ \ \Rightarrow \ \ \left \{ {{x\in(-\infty \ ; \ 0)}\cup(4 \ ; \ +\infty) \atop {x > -6}} \right. \ \ \Rightarrow\\\\\\\Rightarrow \ \ x\in(-6 \ , \ 0)\cup(4 \ ; \ +\infty)\\\\\\\lg(x^{2} -4x)=\lg(x+6)\\\\x^{2} -4x=x+6\\\\x^{2} -5x-6=0\\\\D=(-5)^{2} -4\cdot(-6)=25+24=49=7^{2} \\\\\\x_{1} =\frac{5-7}{2} =-1[/tex]

[tex]\displaystyle\bf\\x_{2} =\frac{5+7}{2} =6\\\\Otvet \ : \ -1 \ ; \ 6[/tex]