Let $a,b,c \in \mathbb Q_2^*$, where $\mathbb Q_2$ denotes the 2-adic field. How to compute the Hilbert symbol $(b^2-4ac,2a)$?
By definition, $(b^2-4ac,2a)=1$, if the equation $z^2-(b^2-4ac)x^2-2ay^2=0$ has a non-trivial solution and $(b^2-4ac,2a)=-1$ otherwise.
I know, that for $\alpha=2^mu$ and $\beta=2^n v$ (with $u,v\in U$) we have $(\alpha,\beta)=(-1)^{\varepsilon (u)\varepsilon (v)+m\omega (v)+n\omega (u)}$.
But I don't think, that it is helpful, because I need to write $b^2-4ac$ as $2^k w$ with $w\in U$.
Are there other ways to compute it?