From Förberedande kurs i matematik 2

If we subtract \displaystyle 2z from both sides,

\displaystyle iz+2-2z=-3

and then subtract \displaystyle 2 from both sides, we have \displaystyle z terms left only on the left-hand side,

\displaystyle iz-2z=-3-2\,\textrm{.}

After taking out a factor \displaystyle z from the left-hand side,

\displaystyle (i-2)z=-5\,,

we obtain, after dividing by \displaystyle -2+i,

\displaystyle \begin{align} z &= \frac{-5}{-2+i} = \frac{-5(-2-i)}{(-2+i)(-2-i)} = \frac{(-5)\cdot(-2)-5\cdot(-i)}{(-2)^2-i^2}\\[5pt] &= \frac{10+5i}{4+1} = \frac{10+5i}{5} = 2+i\,\textrm{.}\end{align}

A quick check shows that \displaystyle z=2+i satisfies the original equation,

\displaystyle \begin{align} \text{LHS} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt] \text{RHS} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.} \end{align}