Quadratic Nimber Equation I'd like to ask how to solve the quadratic nimber equation $x\otimes x \oplus b \otimes x \oplus c=0$, where $\otimes$ is nim multiplication and $\oplus$ is nim addition.

From now on, all arithmetic is nimber arithmetic, except $2^i$ will refer to the usual power of two. When $b=0$, the unique solution is $x=\sqrt{c}$, so assume $b\neq 0$.
In order to solve $z^2+bz+d=0$, use the substitution $z=bx$ to reduce this to 
$$
x^2+x=c.
$$
where $c=d/b^2$. That is, letting $f(x)=x^2+x$, you need $f^{-1}(c)$. 
Note that $f$ is a linear map whose kernel is $\{0,1\}$, which means $f$ is a bijection from the subspace of even nimbers to all nimbers. I will use $f^{-1}(y)$ to denote the unique even nimber $x$ for which $f(x)=y$. Note $f^{-1}$ is also a linear map.
More specifically, it can be shown that 

Prop: For any $n\ge 1$, $f$ is a surjective map from the interval $[2^n,2^{n+1})$ to $[2^{n-1},2^n)$. 

This is proven in Siegel, Combinatorial Game Theory, p. 217 (Lemma 5.6).
Writing $c=\sum_i b_i 2^i$ with $b_i\in \{0,1\}$, we have $f^{-1}(c) = \sum_i b_i f^{-1}(2^i)$, so it suffices to compute $f^{-1}(2^i)$ for any $i$. This can be done recursively as follows. By the above proposition, there are (easily computable) numbers $a_{ij}\in \{0,1\}$ for which
$$
f(2^{i+1})=2^i + \sum_{j=0}^{i-1} a_{ij}2^j
$$
Therefore,
$$
f^{-1}(2^i) = 2^{i+1}+\sum_{j=0}^{i-1}a_{ij}f^{-1}(2^i)
$$
so $f^{-1}(2^i)$ can be computed from previously computed values $f^{-1}(2^j)$. 
