Let $F$ be a field of characteristic not $2$, and let $K$ be an extension of $F$ with $[K: F] = 2$. Show that $K = F (\sqrt{a})$ for some $a\in F$

Pick $\beta \in K\setminus F$. Then $1,\beta,\beta^2$ are $F$-linear dependent, but we know that $1,\beta$ are not. Hence $\beta^2+p\beta+q=0$ with $p,q\in F$. Let $\alpha=\beta+\frac12\cdot p$ (this is where the characteristic is used: otherwise we are not allowed to divide by $2$). Now verify that $\alpha^2\in F$.

Consider the minimal polynomial $x^2+ax+b$, then translate by $a/2$. We are allowed to do so since the characteristic is not 2, then the minimal polynomial becomes $x^2-d$, hence $F=K(\sqrt{d})$ as wanted.