$\alpha = \frac{1}{2}(1+\sqrt{-19})$ I have asked a similar question here before, which was about ring theory, but it is slightly different today and very trivial. 
$\alpha = \frac{1}{2}(1+\sqrt{-19})$ 
Here, $\alpha$ is a root of $\alpha^2 - \alpha + 5$ and $\alpha^2 = \alpha - 5$, but I can't seem to understand this. Could anyone please explain how $\alpha$ is a root of $\alpha^2 - \alpha + 5$?

$\alpha^2-\alpha+5=\frac{1}{4}(1+\sqrt{-19})^2-\frac{1}{2}(1+\sqrt{-19})+5=\frac{1}{4}(1+2\sqrt{-19}+(-19))-\frac{1}{4}(2+2\sqrt{-19})+\frac{1}{4}\cdot 20=\frac{1}{4}(1+2\sqrt{-19}+(-19)-2-2\sqrt{-19}+20)=0$
