Solve the following quadratic equations by factorization:

We have been given

$\frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}$

$7\left(x^{2}+9-6 x-x^{2}-9-6 x\right)=48\left(x^{2}-9\right)$

$48 x^{2}+84 x-432=0$

$4 x^{2}+7 x-36=0$

Therefore,

$4 x^{2}+16 x-9 x-36=0$

$4 x(x+4)-9(x+4)=0$

$(4 x-9)(x+4)=0$

Therefore,

$4 x-9=0$

$4 x=9$

$x=\frac{9}{4}$

or,

$x+4=0$

$x=-4$

Hence, $x=\frac{9}{4}$ or $x=-4$.