Roots of equation $x^n\left(2-x\right)^{2}=a$?

The first thing you can do is to study the trivial case $a=0$ which gives : $$x^n(2-x)^2=0$$ The roots are obviously $x=0$ with order $n$ and $x=2$ with order $2$. You can try to solve other cases to sketch the general form of the solution if it exists.

You can also try some particular values of $n$ :

For $n=0$ you have :

$$x=2\pm\sqrt{a}$$

Which is an order $2$ root.

For $n=1$ I tried with Wolfram Alpha which provides only one ugly solution, which has to be order $3$

For $n=2$ you have : $$x^2(2-x)^2=a$$ $$x(2-x)=2x-x^2=\pm\sqrt{a}$$ Hence the solutions are : $$x=1\pm\sqrt{1\pm\sqrt{a}}$$ Depending on $a$ you can have up to four different solutions with order $1$ each.

It seems complicated to find a general expression as you can have different number/orders of solutions depending on the choice of $(a,n)$

See the following link (theorem 2) https://www.researchgate.net/publication/262973394_Solution_of_Polynomial_Equations_with_Nested_Radicals

He solves :

$$Aqx^{p}+x^{q}=1$$

Your equation is :

$$2x^m-x^{m+1}-\sqrt{a}=0$$ If you make the following substitution :

$x=y\beta$

We have :

$$2\beta^{m}y^m-\beta^{m+1}y^{m+1}-\sqrt{a}=0$$

So divide by $-\beta^{m+1}$ you have:

$$\frac{-2}{\beta}y^m+y^{m+1}+\frac{\sqrt{a}}{\beta^{m+1}}=0$$

And make the last substitutions :

$$\beta^{1}=\frac{1}{m+1}$$ and $$\sqrt{a}=\beta^{m+1}$$

Finally we obtain :

$$(-2)(m+1)y^m+y^{m+1}+1=0$$ So you can apply the theorem 2 with $A=-2$ and $q=m+1$ Another way is to use the theorem 1 with $b=0$ but we obtain a nested radical... Ps:

It's a partial solution because there is some conditions on $\sqrt{a}$... If you want other details see this Solving 5th degree or higher equations