Let #u=x-1#. We can then rewrite the left hand side of the equation as

#sqrt(u+4-4sqrt(u)) + sqrt(u+9-6sqrt(u))#

#=sqrt((sqrt(u)-2)^2) + sqrt((sqrt(u)-3)^2)#

#=|sqrt(u)-2| + |sqrt(u)-3|#

Note the presence of #sqrt(u)# in the equation and that we are only looking for real values, so we have the restriction #u >= 0#. With that, we will now consider all remaining cases:

**Case 1:** #0 <= u <= 4#

#|sqrt(u)-2| + |sqrt(u)-3| = 1#

#=> 2-sqrt(u) + 3-sqrt(2) = 1#

#=> -2sqrt(u) = -4#

#=> sqrt(u) = 2#

#=> u = 4#

Thus #u=4# is the only solution in the interval #[0, 4]#

**Case 2:** #4 <= u <= 9#

#|sqrt(u)-2| + |sqrt(u)-3| = 1#

#=> sqrt(u)-2 + 3 - sqrt(u) = 1#

#=> 1=1#

As this is a tautology, every value in #[4, 9]# is a solution.

**Case 3:** #u >= 9#

#|sqrt(u)-2| + |sqrt(u)-3| = 1#

#=> sqrt(u) - 2 + sqrt(u) - 3 = 1#

#=> 2sqrt(u) = 6#

#=> sqrt(u) = 3#

#=> u = 9#

Thus #u = 9# is the only solution in the interval #[9, oo)#

Taken together, we have #[4, 9]# as the solution set for real values of #u#. Substituting in #x = u+1#, we arrive at the final solution set #x in [5, 10]#

Looking at the graph of the left hand side, this matches with what we would expect: