Solve each of the following quadratic equations:

We write, $-2 x=-3 x+x$ as $3 x^{2} \times(-1)=-3 x^{2}=(-3 x) \times x$

$\therefore 3 x^{2}-2 x-1=0$

$\Rightarrow 3 x^{2}-3 x+x-1=0$

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

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

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

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

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