Question:
$a b x^{2}+\left(b^{2}-a c\right) x-b c=0$
Solution:
Given:
$a b x^{2}+\left(b^{2}-a c\right) x-b c=0$
$\Rightarrow a b x^{2}+b^{2} x-a c x-b c=0$
$\Rightarrow b x(a x+b)-c(a x+b)=0$
$\Rightarrow(b x-c)(a x+b)=0$
$\Rightarrow b x-c=0$ or $a x+b=0$
$\Rightarrow x=\frac{c}{b}$ or $x=\frac{-b}{a}$
Hence, the roots of the equation are $\frac{c}{b}$ and $\frac{-b}{a}$.