$\displaystyle (3-i)(z+4-2i)=10+20i$
My attempt:
$\displaystyle \begin{array}{rcrcrc}
(3-i)(z+4-2i)=10+20i\\
3z + 12-6i-zi-4i+2i^{2}=10+20i\\
3z+10-10i-zi=10+20i\\
3z-zi=30i
\end{array}$
I have no idea where to go from there.
$\displaystyle (3-i)(z+4-2i)=10+20i$
My attempt:
$\displaystyle \begin{array}{rcrcrc}
(3-i)(z+4-2i)=10+20i\\
3z + 12-6i-zi-4i+2i^{2}=10+20i\\
3z+10-10i-zi=10+20i\\
3z-zi=30i
\end{array}$
I have no idea where to go from there.
You are left with 2 equations that you need to solve for x,y:
$\displaystyle 3x+y = 0 \Rightarrow x = \frac{-y}{3}$
$\displaystyle (3y-x)i = (30)i \Rightarrow 3y - x = 30$
Find the possible values for x,y then substitute them into $\displaystyle z = x + yi$ and get your values of z.
Spoiler: