x+√x -2 < [x] +[√x] ≤ x+√x
x+√x -2 < 2x-2 ≤ x+√x
therefore x+√x -2 < 2x-2 →x>1
and 2x-2 ≤ x+√x → x≤4
now lhs is integer therefore rhs should also be integer. hence x is integer.
form the above bound the only integers in range are 2,3,4
plugging them into the eqn we get x=3 and x=4 as the solutions.