Completing the square but in different situation 
Solve the equation $$x^2+4\left(\frac{x}{x-2}\right)^2=45$$

My attempt,
I decided to use completing the square method, so I change it to $$x^2+\left(\frac{2x}{x-2}\right)^2=45$$
But I never encounter this before. Normally, for example $x^2+4x=5$, we can change it to $x^2+4x+(\frac{4}{2})^2=45+(\frac{4}{2})^2$. But in this question is different. Could someone give me some hints for it? Thanks in advance.
 A: The suggested solution posted by OP provide a clever method.
\begin{align}
x^2+4\left(\frac{x}{x-2}\right)^2&=45\\
x^2+2(x)\left(\frac{2x}{x-2}\right)+\left(\frac{2x}{x-2}\right)^2&=45+4\left(\frac{x^2}{x-2}\right)\\
\left(x+\frac{2x}{x-2}\right)^2&=45+4\left(\frac{x^2}{x-2}\right)\\
\left(\frac{x^2}{x-2}\right)^2&=45+4\left(\frac{x^2}{x-2}\right)\\
\left(\frac{x^2}{x-2}\right)^2-4\left(\frac{x^2}{x-2}\right)+4&=49\\
\left(\frac{x^2}{x-2}-2\right)^2&=49\\
\end{align}
A: Well, we have:
$$x^2+4\cdot\left(\frac{x}{x-2}\right)^2=45\tag1$$
Bring together using a common denominator:
$$\frac{x^2\cdot\left(x^2-4x+8\right)}{\left(x-2\right)^2}=45\tag2$$
Multiply both sides by $\left(x-2\right)^2$:
$$x^2\cdot\left(x^2-4x+8\right)=45\cdot\left(x-2\right)^2\tag3$$
Expand out terms of the right hand side:
$$x^2\cdot\left(x^2-4x+8\right)=45x^2-180x+180\tag4$$
Subtract $45x^2-180x+180$ from both sides:
$$x^2\cdot\left(x^2-4x+8\right)-45x^2+180x-180=0\tag5$$
The left hand side factors into a product with three terms:
$$\left(x-6\right)\cdot\left(x-3\right)\cdot\left(x^2+5x-10\right)=0\tag6$$
