Let $K$ be a field with characteristic different from two, then if a extension of fields $L/K$ is such that $[L:K]=2$ then there is a element $\alpha\in L$ such that $L=K(\alpha)$ and $\alpha^2\in K$.
My attempt: If $[L:K]=2$ then I take the basis $B=\{1,\alpha\}$ with $ \alpha\in L $. I know that the set $C=\{1,\alpha,\alpha^2\}$ is linearly dependent, otherwise we have a set with three linearly independent elements contradicting the fact that $[L:K]=2$. Then there are $c,d \in K$ not all zero, such that $c+d\alpha+\alpha^2=0$. Hence $\alpha$ is a zero of the polynomial $p(x)=c+dx+x^2 \in K[x]$ and therefore $F=K(\alpha)$. And $\alpha^2 \in K$ because otherwise $C=\{1,\alpha,\alpha^2\}$ would be linearly dependent.
There is a point in this proof that is not clear. Is it really true that $C=\{1,\alpha,\alpha^2\}$ is linearly independent if $\alpha^2 \in L$? I think that this statement depends on the fact that the characteristic of $K$ is different from two, but i'm not sure.
$K(\alpha) = L$ is true for all $\alpha \in L\setminus K$: it can't be $K$ since $\alpha \notin K$, and it can't be larger than $L$ since both $1$ and $\alpha$ are in $L$. And it can't be an intermediate field between $K$ and $L$ either, because there is no such field (due to the fact that $[L:K] = 2$).
$\alpha^2 \in K$, however, is not true for all $\alpha$, so you have to find something more clever. For an arbitrary $\alpha \in L\setminus K$, you only know that $\alpha^2 = u\alpha+v$, so $\alpha^2-u\alpha = v \in K$. But then $$\left(\alpha-\frac{u}{2}\right)^2 = \alpha^2-u\alpha+\frac{u^2}{4} = v+\frac{u^2}{4} \in K.$$ This is where the fact that the characteristic of $K$ is not $2$ comes in: you can't divide by $2$ in characteristic $2$.