We can eliminate one level of roots with the substitution $u=\sqrt{x-1}$. This of course implies that $x=u^2+1$. Plugging these into the original equation gives

\[\sqrt{u^2-4u+4} + \sqrt{u^2-6u+9}=1. \]

We can then rewrite this as the following

\[\sqrt{(u-2)^2} + \sqrt{(u-3)^2} = 1. \]

It can be seen by inspection u = 3 or u = 2 satisfies this equation. Solving for x then gives x = 10 or x = 5.