$$\begin{aligned} & x^2-(3-2 i) x-(2 i-2)=0 \\ & x=\frac{(3-2 i) \pm \sqrt{(3-2 i)^2-4(1)(-(2 i-2))}}{2(1)} \\ & ==\frac{(3-2 i) \pm \sqrt{9-4-12 i+8 i-8}}{2} \\ & ==\frac{3-2 i \pm \sqrt{-3-4 i}}{2} \\ & =\frac{3-2 i \pm \sqrt{(1)^2+(2 i)^2-2(1)(2 i)}}{2} \\ & =\frac{3-2 \mathrm{i} \pm(1-2 \mathrm{i})}{2} \\ & \Rightarrow \frac{3-2 \mathrm{i}+1-2 \mathrm{i}}{2} \text { or } \frac{3-2 \mathrm{i}-1+2 \mathrm{i}}{2} \\ & \Rightarrow 2-2 \mathrm{i} \text { or } 1+0 \mathrm{i} \\ & \text { So } \alpha \gamma+\beta \delta=2(1)+(-2)(0)=2 \end{aligned}$$