= Solution
The notation $K=F(\delta)$ means that $K$ is the smallest <subfield> containing both $F$ and $\delta$, namely the <simple field extension> generated by $\delta$. Its elements can be expressed as rational functions in $\delta$ with coefficients in $F$ and nonzero denominator.
Write $d=\delta^2\in F$. The element $\delta$ is a root of $X^2-d$, so its <minimal polynomial> over $F$ has degree at most two. Since $\delta\notin F$, that degree cannot be one. The degree of the <simple field extension> equals the degree of the <minimal polynomial>, giving
$$
\boxed{[F(\delta):F]=2.}
$$
In particular $1,\delta$ form an $F$-basis, so every element is uniquely $a+b\delta$. This argument also works in characteristic two; separability is not needed for this part.
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