What do I do if I have to solve the usual quadratic equation $X^2+bX+c=0$ where $b,c$ are in a field of characteristic 2? As pointed in the comments, it can be reduced to $X^2+X+c=0$ with $c\neq 0$.


Usual completion of square breaks. For a finite field there is [Chen Formula](http://www.ams.org/mathscinet-getitem?mr=680147) that roughly looks like $X=\sum\_{m} c^{4^m}$. I am more interested in the local field $F((z))$ or actually an arbitrary field of characteristic 2. 



How to solve a quadratic equation in characteristic 2 ?
I think this solves $X^2+X+c=0$ over $F((t))$:


I want to assume that $c\in F[[t]]$. If not, say $c=at^{-m}+...$, then the quadratic has no solutions when $m$ is odd or $a$ is not a square, and otherwise the substitution $X\mapsto X+\sqrt{a}t^{-m/2}$ gives a new equation with smaller $m$. So, after finitely many steps $c=c\_0+c\_1t+...$ is integral.


Because $X^2+X+c$ has derivative $1$, by Hensel's lemma the equation has a solution if and only the constant term $c\_0$ is of the form $d^2+d$ for some $d$ in $F$. And if it is, Hensel's approximations are obtained by starting with an approximate solution $x\_0=d$ and recursively computing $x\_{m+1}=x\_m-f(x\_m)/f'(x\_m)=x\_m^2+c$. This gives
$$
 x = d + \sum\_{n=0}^\infty (c-c\_0)^{2^n}
$$
as the solution (the partial sums are the $x\_m$). Actually, the approach seems to work over any complete field, reducing the problem to the residue field. Hope this helps. 


