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.