If α+i β and γ+i δ are the roots of x^2-(3-2 i) x-(2 i-2)=0, i=√-1, then α γ+β δ is equal to :

$$\begin{aligned} & x^2-(3-2 \mathrm{i}) \mathrm{x}-(2 \mathrm{i}-2)=0 \\ & \mathrm{x}=\frac{(3-2 \mathrm{i}) \pm \sqrt{(3-2 \mathrm{i})^2-4(1)(-(2 \mathrm{i}-2))}}{2(1)} \\ & ==\frac{(3-2 \mathrm{i}) \pm \sqrt{9-4-12 \mathrm{i}+8 \mathrm{i}-8}}{2} \\ & ==\frac{3-2 \mathrm{i} \pm \sqrt{-3-4 \mathrm{i}}}{2} \\ & =\frac{3-2 \mathrm{i} \pm \sqrt{(1)^2+(2 \mathrm{i})^2-2(1)(2 \mathrm{i})}}{2} \\ & =\frac{3-2 \mathrm{i} \pm(1-2 \mathrm{i})}{2} \\ & \Rightarrow \frac{3-2 \mathrm{i}+1-2 \mathrm{i}}{2} \text { or } \frac{3-2 \mathrm{i}-1+2 \mathrm{i}}{2} \\ & \Rightarrow 2-2 \mathrm{i} \text { or } 1+0 \mathrm{i} \\ & \text { So } \alpha \gamma+\beta \delta=2(1)+(-2)(0)=2\end{aligned}$$​x2−(3−2i)x−(2i−2)=0x=2(1)(3−2i)±(3−2i)2−4(1)(−(2i−2))​​==2(3−2i)±9−4−12i+8i−8​​==23−2i±−3−4i​​=23−2i±(1)2+(2i)2−2(1)(2i)​​=23−2i±(1−2i)​⇒23−2i+1−2i​ or 23−2i−1+2i​⇒2−2i or 1+0i So αγ+βδ=2(1)+(−2)(0)=2​