Решите уравнение 1/x-1+1/x^2-1=5/8

ОДЗ: х=1,-1.
$\frac{1}{x-1}+ \frac{1}{x^2-1}= \frac58\\ \frac{1}{x-1}+ \frac{1}{(x-1)(x+1)}= \frac58\\8(x-1)(x+1)* \frac{1}{x-1}+8(x-1)(x+1)* \frac{1}{(x-1)(x+1)}=8(x-1)(x+1)*$ $\frac58\\8(x+1)+8=5(x-1)(x+1)\\8x+8+8=5(x^2-1)\\8x+16=5x^2-5\\5x^2-8x-21=0\\D_1=( \frac{b}{2})^2-ac=16+105=121\\x_1= \frac{- \frac{b}{2}+ \sqrt{D}}{a}= \frac{4+11}{5}=3\\x_2=\frac{- \frac{b}{2}- \sqrt{D}}{a}= \frac{4-11}{5}=-1,4$