Solution
$\mathrm{x}^{2}-(3-2 \mathrm{i}) \mathrm{x}-(2 \mathrm{i}-2)=0$
\[
\begin{aligned}
& x=\frac{(3-2 i) \pm \sqrt{(3-2 i)^{2}-4(1)(-(2 i-2))}}{2(1)}=\frac{3-2 i \pm \sqrt{-3-4 i}}{2} \\
& x=\frac{3-2 i \pm \sqrt{(1)^{2}+(2 i)^{2}-2(1)(2 i)}}{2}=\frac{3-2 i \pm(1-2 i)}{2} \Rightarrow x=2-2 i \text { or } 1+0 i
\end{aligned}
\]
According to question $\alpha+\beta \mathrm{i}=2-2 \mathrm{i}$ and $\gamma+\delta \mathrm{i}=1+\mathrm{i} .0$ we get $\alpha=2, \beta=-2, \gamma=1, \delta=0$ So, $\alpha \gamma+\beta \delta=2(1)+(-2)(0)=2$
.
