Primavera stated a question, and mentioned that...
There are only three quadratic extensions over $\mathbb{Q}$ in $\mathbb{Q}(\zeta_{20})$ as $\mathbb{Q}(\sqrt{5}),\mathbb{Q}(i),\mathbb{Q}(\sqrt{5}i)$, and every nontrivial intermediate field between $\mathbb{Q}$ and $\mathbb{Q}(\zeta_{20})$ is normal over $\mathbb{Q}$ since $\text{Gal}(\mathbb{Q}(\zeta_{20})/\mathbb{Q})$ is abelian.
I assume that $\mathbb{Q}(\zeta_{20})$ is the field generated by all roots, which orders are exactly 20.
Why is it true that $\mathbb{Q}(\sqrt{5})$ lies inside $\mathbb{Q}(\zeta_{20})$?
Which minimal polynomial does $\sqrt{5}$ have, $x^2 - 5$ doesn't appear to lie in $\mathbb{Q}(\zeta_{20})$?
$\zeta_{20}$ is the first $20$th root of unity, so it follows that $\zeta_{20}^4=\zeta_5$, the first fifth root of unity. Now $$\zeta_5+\zeta_5^4=2\cos\frac{2\pi}5=\frac{\sqrt5-1}2$$ so this shows $\sqrt5$, and hence $\mathbb Q(\sqrt5)$, lies in $\mathbb Q(\zeta_{20})$.