Помогите решить пожалуйста уравнение

[latex] \frac{-2(20-x-x^2)}{(5+x)^2}+ \frac{4x-12}{(x-3)(5+x)}=3 [/latex] ОДЗ: x≠3;  x≠ -5 [latex] \frac{2(x^2+x-20)}{(5+x)^2} + \frac{4(x-3)}{(x-3)(5+x)}-3=0 \\ \\ \frac{2(x^2+5x-4x-20)}{(5+x)^2}+ \frac{4}{5+x}- \frac{3(5+x)}{5+x}=0 \\ \\ \frac{2(x(x+5)-4(x+5))}{(5+x)^2}+ \frac{4}{5+x}- \frac{15+3x}{5+x}=0 \\ \\ \frac{2(x+5)(x-4)}{(5+x)^2}+ \frac{4}{5+x}- \frac{15+3x}{5+x}=0 \\ \\ \frac{2(x-4)}{5+x}+ \frac{4}{5+x}- \frac{15+3x}{5+x}=0 \\ \\ 2x-8+4-15-3x=0 \\ -x-19=0 \\ -x=19 \\ x=-19 [/latex] Ответ: -19.