Solve each of the following quadratic equations:

We write, $-3 x=3 x-6 x$ as $9 x^{2} \times(-2)=-18 x^{2}=3 x \times(-6 x)$

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

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

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

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

$\Rightarrow 3 x+1=0$ or $3 x-2=0$

$\Rightarrow x=-\frac{1}{3}$ or $x=\frac{2}{3}$

Hence, the roots of the given equation are $-\frac{1}{3}$ and $\frac{2}{3}$.