If $[K:\Bbb{Q}]=2$ then $K=\Bbb{Q}(\sqrt{d})$.

Question

I am stuck on one question and sincerely have no idea how to proceed. Let $K$ be a field containing $\Bbb{Q}$ such that $[K : \Bbb{Q} ] = 2$. Prove that there exists a square free integer $d$ such that $K=\Bbb{Q}[\sqrt{d}]$. Any hint to proceed.

Answer 1

HINT 1: If $x\in K-\Bbb{Q}$ then $1$, $x$ and $x^2$ are linearly dependent over $\Bbb{Q}$ because $[K:\Bbb{Q}]=2$.

HINT 2: This implies there exist $a,b,c\in\Bbb{Q}$ such that $ax^2+bx+c=0$, and so $$(2ax+b)^2=b^2-4ac.$$