let's keep in mind that i² = -1, and let's use the conjugate of the denominator.
[tex]\cfrac{2+5i}{3-2i}\cdot \cfrac{3+2i}{3+2i}\implies \cfrac{(2+5i)(3+2i)}{\underset{\textit{difference of squares}}{(3-2i)(3+2i)}}\implies \cfrac{6+4i+15i+10i^2}{(3)^2~~ - ~~(2i)^2} \\\\\\ \cfrac{6+19i+10(\stackrel{i^2}{-1})}{9~~ - ~~(2^2 i^2)}\implies \cfrac{6+19i-10}{9~~ - ~~4(\underset{i^2}{-1})}\implies \cfrac{-4+19i}{9+4}\implies -\cfrac{4}{13}+\cfrac{19i}{13}[/tex]
