Solving a factorized quadratic matrix equation

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I have an equation such as

$$ X^{T}AX + X^{T}B + C = 0 $$

where $X, A, B,$ and $C$ are members of $\mathbb{R}^{n\times n}$. I would like to solve for $X.$ Nothing can be said about the symmetry or invertibility of $A, B,$ or $C$. In fact, $A$ is definitely not invertible.

$X$ has elements $\in \{0, 1\}$ on the diagonal only, so $X = X^{T}$ and $X = X^{n}, \forall n \gt 0$ so if I can get the quadratic terms together, it becomes a linear equation.

I thought to complete the square to get it in the form $$ (X - H)^{T}A(X - H) + K = 0 $$

where $H = -(A + A^{T})^{-1}B$ and $K = C - H^{T}AH$ but this didn't work because $(A + A^{T})$ is not invertible.

Obviously for real numbers, one could use the quadratic formula but I'm not sure how to proceed for the matrix case. I checked other posts for a solution but could not find anything for this type of problem specifically. This post seemed relevant but not relevant enough for me to solve my problem.

EDIT: Added that factorization didn't work, $A$ is not invertible, and $X = X^{2}$

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If $X=X^2$ then

$$X^TAXX+X^TBX+CX = X^T(A+B)X+CX = 0$$

and

$$X^TX^T(A+B)X+X^TCX = X^T(A+B+C)X = 0$$

Can you proceed from here?


$\textbf{Edit:}$ What can also help is that $X$ can be written as

$$X = \sum_i c_i \cdot e_ie_i^T$$

where $e_i$ are the standard basis vectors for $\Bbb{R^n}$