How to solve quadratic matrix equations of the form $A^T B A=C$? I want to solve the following matrix equation
$$A^T  
\begin{pmatrix}
 10 & 0 & 0 \\
 0 & 20 & 0 \\
 0 & 0 & 25 \\
\end{pmatrix} A =
\begin{pmatrix}
 \frac{35}{2} & \frac{5 \sqrt{3}}{2} & 0 \\
 \frac{5 \sqrt{3}}{2} & \frac{25}{2} & 0 \\
 0 & 0 & 25 \\
\end{pmatrix}$$
In the above formula, $A^T$ is the transpose of matrix A.
At present, I don't have a good way. I only know that the reference answer of matrix A is 
$$\left(\begin{array}{ccc}
 \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\
 \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\
 0 & 0 & 1 \end{array}\right)$$

Call the matrix $$C'=\begin{bmatrix}\frac{35}{2}&\frac{5\sqrt 3}{2}\\\frac{5\sqrt 3}{2}&\frac{25}{2}\end{bmatrix}=\begin{bmatrix}a&b\\b&d\end{bmatrix}$$ Apply the 'diagonalization of a symmetric matrix by congruence' formula (which comes from 'completing the square' in the corresponding quadratic form). If $$E=\begin{bmatrix}1&-\frac{b}{a}\\0&1\end{bmatrix}$$ then $$E^TC'E=\begin{bmatrix}a&0\\0&\frac{\det C'}{a}\end{bmatrix}.$$ Thus $$E=\begin{bmatrix}1&\frac{-\sqrt 3}{7}\\0&1\end{bmatrix}$$
and $$E^TC'E=\begin{bmatrix}\frac{35}{2}&0\\0&\frac{80}{7}\end{bmatrix}.$$ Let
 $$F=\begin{bmatrix}\frac{2}{\sqrt 7}&0\\0&\frac{\sqrt 7}{2}\end{bmatrix}$$ Then 
$$F^TE^TC'EF=\begin{bmatrix}10&0\\0&20\end{bmatrix}.$$ Let G be the 3 x 3 matrix whose upper left 2 x 2 block is $EF$ and whose element in row 3 and column 3 is 1 and whose other elements are 0. Then $$G^TCG=\begin{bmatrix}10&0&0\\0&20&0\\0&0&25\end{bmatrix}.$$ Thus $A=G^{-1}$ 
