@TedShifrin Since $[F:K]=2$ the degree of $Irr(a,K)$ should divide $2$. So, the degree is $1$ or $2$.
If the degree is $1$, $Irr(a,K)=x-a$. The solution is $a \in F$
If the degree is $2$, $Irr(a,K)=x^2+Ax+B, A,B \in K$. The one solution is $a \in F$ and the other one is $-a-A \in F$ ($a \in F$ and $A \in K \Rightarrow A \in F$)