Find all real and imaginary solutions to the equation. $$ \left(x^{2}-6 x\right)^{2}-35\left(x^{2}-6 x\right)-200=0 $$

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Solve the equation $$ (x^{2}-6x)^{2}-35(x^{2}-6x)-200=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\left(x^{2}-6xight)^{2}-35\left(x^{2}-6xight)-200=0$$
- step1: Calculate:
  $$x^{4}-12x^{3}+x^{2}+210x-200=0$$
- step2: Factor the expression:
  $$\left(x-10ight)\left(x-5ight)\left(x-1ight)\left(x+4ight)=0$$
- step3: Separate into possible cases:
  $$\begin{align}&x-10=0\&x-5=0\&x-1=0\&x+4=0\end{align}$$
- step4: Solve the equation:
  $$\begin{align}&x=10\&x=5\&x=1\&x=-4\end{align}$$
- step5: Rewrite:
  $$x_{1}=-4,x_{2}=1,x_{3}=5,x_{4}=10$$
The real solutions to the equation are $$ x = -4, 1, 5, 10 $$.
The real solutions are $$ x = -4, 1, 5, 10 $$.

Find all real and imaginary solutions to the equation.

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