Решить уравнение 4/(x+2)-3/(x-2)-12/(4-x^2)=1/7

[latex] \frac{4}{x+2}- \frac{3}{x-2} - \frac{12}{4-x^2} = \frac{1}{7y} |(4-x^2)=(2-x)(2+x)\\ \frac{4(2-x)+3(2+x)-12-1(4-x^2)}{4-x^2} =0 \\ [/latex] [latex]4(2-x)+3(2+x)-12-1(4-x^2)=0 \\ 8-4x+6+3x-12-4+4x^2=0 \\ 4x^2-x-2=0 \\ D=1+32=33 \\ x1= \frac{1- \sqrt{33} }{8} \\ x2= \frac{1+ \sqrt{33} }{8}[/latex]