$\begin{array}{l}
3 x^{2}-x-2=0 \\
\Rightarrow 9 x^{2}-3 x-6=0 \quad \text { (Multiplying both sides by 3) } \\
\Rightarrow 9 x^{2}-3 x=6 \\
\Rightarrow(3 x)^{2}-2 \times 3 x \times \frac{1}{2}+\left(\frac{1}{2}\right)^{2}=6+\left(\frac{1}{2}\right)^{2} \quad \quad\left[\text { Adding }\left(\frac{1}{2}\right)^{2}\right. \text { on both sides] }
\end{array}$
$\begin{array}{l}
\Rightarrow\left(3 x-\frac{1}{2}\right)^{2}=6+\frac{1}{4}=\frac{25}{4}=\left(\frac{5}{2}\right)^{2} \\
\Rightarrow 3 x-\frac{1}{2}=\pm \frac{5}{2} \quad \text { (Taking square root on both sides) } \\
\Rightarrow 3 x-\frac{1}{2}=\frac{5}{2} \text { or } 3 x-\frac{1}{2}=-\frac{5}{2} \\
\Rightarrow 3 x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3 \text { or } 3 x=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2 \\
\Rightarrow x=1 \text { or } x=-\frac{2}{3}
\end{array}$

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