Recognise that
$\displaystyle 6a^2+a-2 = (3a+2)(2a-1)$ and
$\displaystyle 2a^2-3a+1 = (2a-1)(a-1)$.
Putting these together, we obtain
$\displaystyle \frac{6a^2 + a - 2}{(2a^2 - 3a + 1} = \frac{(3a+2)(2a-1)}{(2a-1)(a-1)}=\frac{3a+2}{a-1}$,
where the last step is done under the assumption that
$\displaystyle a\ne\frac{1}{2}_{.}$
If we do not have this assumption, we cannot do the cancellation.