Notice that the insider parts of both terms are perfect squares. Thus this equation can be simplified to: $$\big|\sqrt{x-1}-2\big| +\big|\sqrt{x-1}-3\big| = 1$$ When absolute values are involved, we need to study their \textit{zero} points in order to remove the absolute value signs. As there are two terms, there will be two \textit{zero} points: $$\sqrt{x-1} - 2 = 0\quad\text{and}\quad\sqrt{x-1}-3=0$$ Do casework: (1) When $\sqrt{x-1} < 2$, the original equation is equivalent to: $$-(\sqrt{x-1}-2)-(\sqrt{x-1}-3)=1\implies \sqrt{x-1}=2$$ This contradicts to the assumption $\sqrt{x-1} < 2$. (2) When $2\le\sqrt{x-1}< 3$, the original equation is equivalent to: $$(\sqrt{x-1}-2)-(\sqrt{x-1}-3)=1$$ This is an identity which implies any $x$ satisfying the assumption will be a solution. Or $$2\le\sqrt{x-1}< 3\implies \boxed{5\le x < 10}$$ (3) When $\sqrt{x-1} \ge 3$, the original equation is equivalent to: $$(\sqrt{x-1}-2)+(\sqrt{x-1}-3)=1 \implies x = \boxed{10}$$ Therefore we conclude all solutions are $\boxed{5\le x \le 10}$.