To prove: $ [K : \mathbb{Q}] = 2 \ \Longrightarrow \ \exists \zeta \text{ primitive root of unity}, \ \mathbb{Q}(\zeta) \ \supseteq \ K $ I have to show that the following statement is true:
Let $K$ Be a field extending $\mathbb{Q}$ such that $[K: \mathbb{Q}] \ = \ 2$. Then there is a root of unity $\zeta$ such that $K \subseteq \mathbb{Q}(\zeta)$.

What I did
Since $2$ is prime, there can't be a field $L$ such that 
$L \supset K \supset \mathbb{Q}$. That means that
$ K \ = \ K(\alpha) \ \cong \ \mathbb{Q}[X] /(X^2 +aX +b) $
for some $a,b \in \mathbb{Q}$. Now I have to show that one always can find a number $n \in \mathbb{N}$, and $a_0, a_1, \cdots , a_n \in \mathbb{Q}$ such that
$$
\alpha = a_0 + a_1\zeta + a_2\zeta^2 + \cdots + a_n-1\zeta^{n-1}
$$
But I don't really know how.

Try to answer my own question using the hint
Since the roots of $X^2 +aX+b$ are $\frac{\pm\sqrt{a^2-4b}-a}{2}$, we only have to add $\sqrt{a^2-4b}$ to get the field $K$. Now we replace $m = a^2-4b$, and we split $m$ in prime numbers.
$$
m \ = \ p_1^{k_1} \cdots p_t^{k_n}
 \quad \text{ so } \quad 
\sqrt{m} \ = \ \sqrt{p_1}^{k_1} \cdots \sqrt{p_t}^{k_n}
$$
Now I write 
$\gamma_p 
\ := \ 
\sum_{a=0}^{p-1} \left( \frac{a}{p}\right) \zeta_p^a
$.
We know that 
$
\gamma_p^2
\ = \  
p \left( \frac{-1}{p}\right) 
$,
thus
$
\gamma_p \ = \ \pm \sqrt{\pm p}
$.
Now we see that
$$
\sqrt{m} \ = \ \pm g_{p_1}^{k_1}\cdots g_{p_t}^{k_t}
\ \in \ \mathbb{Q}(\zeta_{p_1}, \cdots \zeta_{p_t})
$$
We know that $\zeta_{p_1} \cdot \zeta_{p_2} \cdots \zeta_{p_t}$ is a primitive root of unity as well. If we link up all this knowledge we have:
$$
\mathbb{Q}(\alpha) \ = \ \mathbb{Q}(\sqrt{m}) \ = \
\mathbb{Q}(\pm g_{p_1}^{k_1}\cdots g_{p_t}^{k_t} ) \ = \ \mathbb{Q}(\zeta_{p_1}, \cdots \zeta_{p_t})
\ = \ \mathbb{Q}(\zeta_{p_1} \cdot \zeta_{p_2} \cdots \zeta_{p_t})
$$
Which means that the product of those primitive roots is the root of unity we had to find.

Hint: As said in the comments, every quadratic extension of $\mathbb{Q}$ is of the form $\mathbb{Q}(\sqrt{\alpha})$ with $\alpha\in\mathbb{Z}$. Try proving the assertion first for $K=\mathbb{Q}(\sqrt{\pm p})$ with $p$ prime. For that, consider the Gauss sum $\gamma = \displaystyle\sum_{a=0}^{p-1} \left(\dfrac{a}{p}\right)\zeta_p^a$ where $\zeta_p$ is a primitive $p$-th root of unity and $\left(\dfrac{a}{p}\right)$ is the Legendre symbol. Prove that $\gamma^2 = p \left(\dfrac{-1}{p}\right)$. Prove that $\gamma$ is fixed by the only  subgroup of $\mathrm{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$ of index $2$.
