Solve the following quadratic equations by factorization:

We have been given

$3 x^{2}-14 x-5=0$

$3 x^{2}-15 x+x-5=0$

$3 x(x-5)+1(x-5)=0$

$(3 x+1)(x-5)=0$

Therefore,

$3 x+1=0$

$3 x=-1$

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

or,

$x-5=0$

$x=5$

Hence, $x=\frac{-1}{3}$ or $x=5$.