Помогите пожалуйста решить уравнение (x/(x+1))^2+((x)/(x-1))^2=90.

Решите задачу:

$\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2=90 \\ \frac{x^2}{(x+1)^2}+\frac{x^2}{(x-1)^2}=90 \\ \frac{x^2(x-1)^2+x^2(x+1)^2}{(x-1)^2(x+1)^2}=90 \\ \frac{x^2((x-1)^2+(x+1)^2)}{(x-1)^2(x+1)^2}=90 \\ \frac{x^2((x^2-2x+1)+(x^2+2x+1))}{(x-1)^2(x+1)^2}=90 \\ \frac{x^2(2x^2+2)}{(x-1)^2(x+1)^2}=90 |(x-1)^2(x+1)^2; x \neq \pm1 \\ x^2(2x^2+2)=90(x-1)^2(x+1)^2 \\ x^2(2x^2+2)=90(x^2-1)^2 \\ 2x^2(x^2+1)=90(x^2-1)^2 ;$ $-88x^4+182x^2-90=0 \\ -2(44x^4-91x^2+45)=0 | :-2 \\ 44x^4-91x^2+45=0 ; t=x^2 \\ 44t^2-91t+45=0 \\ D = 91^2-44*4*45=361 \\ t_{1,2} = \frac{91\pm19}{2} \\ t_1 = \frac{110}{88} \\ t_2 = \frac{72}{88} \\ x_{1,2} = \pm\sqrt{\frac{110}{88}} \\ x_{3,4} = \pm\sqrt{\frac{72}{88}} \\$