We solve the quadratic equation by completing the square: $$\begin{aligned}z^2-2\, z+2=0 &\qquad \color{blue}{\text{the given equation}}\\ \\ \left(z-1\right)^2+1=0 &\qquad\color{blue}{\text{completing the square}}\\ \\ (z-1)^2=-1 &\qquad \color{blue}{\text{constant to the right}} \\ \\ z-1=\,\mathrm{i}\quad \lor\quad z-1=-\,\mathrm{i} &\qquad \color{blue}{\mathrm{i}^2=-1} \\ \\ z=1+\,\mathrm{i}\quad \lor\quad z= 1-\,\mathrm{i}&\qquad \color{blue}{\text{solutions in standard form}} \end{aligned}$$ Above, we have used the logical "or" operator $\lor$.
So, there are two solutions: $$z=1+\,\mathrm{i}\quad\text{or}\quad z=1-\,\mathrm{i}$$

We solve the quadratic equation by completing the square: $$\begin{aligned}z^2-8\, \ii\, z-20=0 &\qquad \color{blue}{\text{the given equation}}\\ \\ \left(z-4\, \ii\right)^2-4=0 &\qquad\color{blue}{\text{completing the square with }\ii^2=-1}\\ \\ (z-4\,\mathrm{i})^2=4 &\qquad\color{blue}{\text{constant to the right}} \\ \\z-4\,\mathrm{i}=2\quad \lor\quad z-4\,\mathrm{i}=-2 &\qquad\color{blue}{\text{taking roots}} \\ \\ z=2+4\,\mathrm{i}\quad \lor\quad z=-2+4\,\mathrm{i} &\qquad\color{blue}{\text{solutions in standard form}}\end{aligned}$$ So there are two solutions: $$z=2+4\,\mathrm{i}\quad\text{or}\quad z=-2+4\,\mathrm{i}$$