Solve the following quadratic equations by factorization:

We have been given,

$9 x^{2}-3 x-2=0$

$9 x^{2}-6 x+3 x-2=0$

$3 x(3 x-2)+1(3 x-2)=0$

$(3 x+1)(3 x-2)=0$

Therefore,

$3 x+1=0$

$3 x=-1$

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

or,

$3 x-2=0$

$3 x=2$

$x=\frac{2}{3}$

Hence, $x=\frac{-1}{3}$ or $x=\frac{2}{3}$.