= Solution
Choose $\alpha\in K\setminus F$. Since $[K:F]=2$, the linearly independent elements $1,\alpha$ form a <basis> of $K$ over $F$. Thus there are $a,b\in F$ with $\alpha^2=a+b\alpha$. The assumption on the <characteristic of a field> allows division by two. Set
$$
\delta=\alpha-\frac b2.
$$
Completing the square gives
$$
\boxed{\delta^2=a+\frac{b^2}{4}\in F.}
$$
Also $\delta\notin F$, since otherwise $\alpha=\delta+b/2$ would lie in $F$. Because $\alpha\in F(\delta)$ and $\delta\in F(\alpha)$, these generated <subfields> agree. The basis representation shows $F(\alpha)=K$, hence
$$
\boxed{K=F(\delta),\qquad\delta^2\in F.}
$$
The square-root generator is obtained by translating an arbitrary generator to remove the linear term of its quadratic relation.
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