If \(\alpha+i \beta\) and \(\gamma+i \delta\) are the roots of \(x^2-(3-2 i) x-(2 i-2)=0, i=\sqrt{-1}\), then \(\alpha \…

\({x}^{2}-\left(3-2i\right)x-\left(2i-2\right)=0\)

\(x=\frac{(3-2i)\pm \sqrt{(3-2i{)}^{2}-4(1)(-(2i-2))}}{2(1)}\)

\(x=\frac{3-2i\pm \sqrt{-3-4i}}{2}\)

\(=\frac{3-2\mathrm{i}\pm \sqrt{(1{)}^{2}+(2\mathrm{i}{)}^{2}-2(1)(2\mathrm{i})}}{2}\)

\(=\frac{3-2\mathrm{i}\pm (1-2\mathrm{i})}{2}\)

\(x=\frac{3-2\mathrm{i}+1-2\mathrm{i}}{2},\frac{3-2i-1+2i}{2}\)

\(x=2–2i,1+0i\)

So \(\alpha \gamma +\beta \delta =2\left(1\right)+\left(-2\right)\left(0\right)=2\)