Solution to a particular quadratic matrix equation I am looking for the solution of the following quadratic matrix equation 
$XAX^{\prime }-XB+A=\mathbf{0}$
where $A$, $B$ and $X$ are square matrices and $B$ is a symmetric positive definite matrix. 
Any suggestion? 
Thanks for your help.  

In the special case where $A$ is Hermitian and $-B\preceq 2A\preceq B$, the equation can be solved as follows. Since $B\succ0$ and $A$ is Hermitian, the two matrices can be simultaneously diagonalised by congruence. More specifically, let $UDU^\ast$ be a unitary diagonalisation of $B^{-1/2}AB^{-1/2}$. (If both $A$ and $B$ are real, $U$ and in turn $P$ can be taken to be real.) Then $B=PP^\ast$ and $A=PDP^\ast$, where $P=B^{1/2}U$. So, if we put $Y=P^{-1}XP$, the equation in question can be rewritten as
$$
YDY^\ast - Y + D = 0.\tag{1}
$$
As $-B\preceq 2A\preceq B$, we have $-I\preceq 2B^{-1/2}AB^{-1/2}\preceq I$, i.e. $-I\preceq 2D\preceq I$. Therefore $|2d_{ii}|\le1$ for each $i$ and we may take $Y$ to be a real diagonal matrix such that
$$
y_{ii}=\begin{cases}
0&\text{ if }d_{ii}=0,\\
\frac{1\pm\sqrt{1-4d_{ii}^2}}{2}&\text{ otherwise}.
\end{cases}
$$
Hence $X=PYP^{-1}$ solves the original equation.
