If we subtract 2z from both sides,
iz+2−2z=−3
and then subtract 2 from both sides, we have z terms left only on the left-hand side,
iz−2z=−3−2
After taking out a factor z from the left-hand side,
(i−2)z=−5
we obtain, after dividing by −2+i,
z=−5−2+i=−5(−2−i)(−2+i)(−2−i)=(−2)2−i2(−5)(−2)−5(−i)=4+110+5i=510+5i=2+i
A quick check shows that z=2+i satisfies the original equation
LHSRHS=iz+2=i(2+i)+2=2i−1+2=1+2i=2z−3=2(2+i)−3=4+2i−3=1+2i