The quadratic rule still applies and

$$\begin{align} z&=\frac{4\pm\sqrt{16-4(4+i2)}}{2}\\\\ &=2\pm \sqrt{-i2}\\\\ &=2\pm \sqrt{2}e^{-i\pi/4}\\\\ &=2\pm(1-i) \end{align}$$

The roots are

$$\bbox[5px,border:2px solid #C0A000]{z=3-i \,\,\,\text{and}\,\,\,z=1+i}$$

This is equivalent to $(z-2)^{2}=-2i=2e^{\frac{i3\pi}{2}}$ so $z-2=\sqrt{2}e^{\frac{i3\pi}{4}}$ or $z-2=-\sqrt{2}e^{\frac{i3\pi}{4}}$.

Notice, we have $$z^2-4z+4+2i=0$$ $$(z-2)^2+2i=0$$ $$(z-2)^2=-2i\iff (z-2)^2=2i^3$$

$$z-2=\sqrt{2}i^{3/2}$$ Since, $i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}$, hence we get $$z-2=\sqrt{2}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^{3/2}=\sqrt{2}\left(\cos\left(2k\pi+\frac{\pi}{2}\right)+i\sin\left(2k\pi+\frac{\pi}{2}\right)\right)^{3/2}$$ $$=\sqrt{2}\left(\cos\left(\frac{4k\pi+\pi}{2}\right)+i\sin\left(\frac{4k\pi+\pi}{2}\right)\right)^{3/2}$$

$$z-2=\sqrt{2}\left(\cos\frac{3(4k\pi+\pi)}{4}+i\sin\frac{3(4k\pi+\pi)}{4}\right)$$

Now, setting $k=0$, we get first root as follows

$$z=\sqrt{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)+2$$ $$=\sqrt{2}\left(-\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}\right)+2$$

$$=-1+i+2=1+i$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{z=1+i}}$$

Now, setting $k=1$, we get second root as follows $$z=\sqrt{2}\left(\cos\frac{15\pi}{4}+i\sin\frac{15\pi}{4}\right)+2$$ $$=\sqrt{2}\left(\frac{1}{\sqrt2}-i\frac{1}{\sqrt2}\right)+2$$

$$=1-i+2=3-i$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{z=3-i}}$$