Solve the following quadratic equations by factorization:

We have been given

$48 x^{2}-13 x-1=0$

$48 x^{2}-16 x+3 x-1=0$

$16 x(3 x-1)+1(3 x-1)=0$

$(16 x+1)(3 x-1)=0$

Therefore,

$16 x+1=0$

$16 x=-1$

$x=\frac{-1}{16}$

or,

$3 x-1=0$

$3 x=1$

$x=\frac{1}{3}$

Hence, $x=\frac{-1}{16}$ or $x=\frac{1}{3}$