Solve the following quadratic equations by factorization:

We have been given

$\frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7}$

$\frac{6}{x^{2}+4 x-5}=\frac{6}{7}$

$x^{2}+4 x-12=0$

$x^{2}+6 x-2 x-12=0$

$x(x+6)-2(x+6)=0$

$(x-2)(x+6)=0$

Therefore,

$x-2=0$

$x=2$

or,

$x+6=0$

$x=-6$

Hence, $x=2$ or $x=-6$