Use $(z-a)(z-b)=z^2-(a+b)+ab=0$ to get $ (a+b)=3 \tag{1} $ $ ab=(3-i) \tag{2} $ From (1) you get $\Im(a)=-\Im(b)$. So $(a_r+ia_i)(b_r-ia_i)=(a_rb_r+a_i^2)+i(-a_r+b_r)a_i=(3-i)$, thus $a_rb_r+a_i^2=3$ and $(b_r-a_r)a_i=-1$. Now you can try a few values like ...
$a_i=1$ and figure out that $a_r=2$ and $b_r=1$. So you finally rewrite it as $(z-(2+i))(z-(1-i))$.
You could also just use the standard way (as proposed by anon) to get: $ \frac{3\pm\sqrt{9-4(3-i)}}{2}=\frac{3\pm\sqrt{-3+4i}}{2}. $ Now note that $(i(2-i))^2=-3+4i$, so $ \frac{3\pm i(2-i)}{2}=\frac{3\pm (2i+1)}{2}=\frac{3\pm 1 }{2} \pm i $