Solve the following quadratic equations by factorization:

We have been given

$a x^{2}+\left(4 a^{2}-3 b\right) x-12 a b=0$

$a x^{2}+4 a^{2} x-3 b x-12 a b=0$

$a x(x+4 a)-3 b(x+4 a)=0$

$(a x-3 b)(x+4 a)=0$

Therefore,

$a x-3 b=0$

$a x=3 b$

$x=\frac{3 b}{a}$

or

$x+4 a=0$

$x=-4 a$

Hence, $x=\frac{3 b}{a}$ or $x=-4 a$.