Answer
$14+32i$
Work Step by Step
We multiply each term of the first complex number with the second complex number and then simplify:
$(4+2i)(6+5i)$
=$4(6+5i)+2i(6+5i)$
=$24+20i+12i+10i^{2}$
=$24+32i+10(-1)$ [We substitute -1 in place of $i^{2}$ as $i^{2}=-1$]
=$24-10+32i$
=$14+32i$
