4/x+2 - 3/x-2 - 12/4-x² = 1/7 решите уравнение

$\frac{4}{x+2}- \frac{3}{x-2} - \frac{12}{4-x^{2} } = \frac{1}{7} \\ \frac{4(x-2)}{x+2}- \frac{3(x+2)}{x-2} - \frac{12}{4-x^{2} }= \frac{1}{7} \\ \frac{4(x-2)-3(x+2)}{(x-2)(x+2)} - \frac{12}{4-x^{2} }= \frac{1}{7} \\ \frac{4x-8-3x-6}{ x^{2} -4} - \frac{12}{4-x^{2} }= \frac{1}{7} \\ \frac{x-14}{ x^{2} -4} - \frac{12}{4-x^{2} }= \frac{1}{7} \\ \frac{x-14-12}{ x^{2} -4} = \frac{1}{7} \\ \frac{x-26}{ x^{2} -4} = \frac{1}{7} \\ (x-26)*7=(x^{2} -4)*1$
7x-182=x^{2} -4
7x-x^{2}=182-4
(x-26)*7=(x^{2} -4)*1
7x-182=x^{2} -4
7x-x^{2}=182-4
7x-x^{2}=178