For all #x# except #x=5# the expression #(x-5)^2# is strictly positive. The value #x=5# is a solution since it sets the left hand side to zero.

Therefore, we can

(1) say that #x=5# is one of the solutions of this inequality and

(2) for all other #x# we can reduce both sides of this inequality by positive #(x-5)^2# without changing the comparison in the inequality getting a simpler inequality to solve:

#(x-1)(x+3) <= 0#

Next, we divide the set of real numbers into intervals by points that set each multiplier to zero (#x=1# and #x=-3#).

Consider #x# as moving from #-oo# on the left to #+oo# on the right.

(a) While its less than #-3#, both #x-1# and #x+3# are negative and, therefore, their product is positive, not good for our inequality.

(b) As soon as #x# crosses #-3#, one of these multipliers, #x+3#, changes sign to positive, while the other, #x-1#, remains negative. Their product is negative, which is a solution.

(c) finally, as #x# crosses #1#, both multipliers are positive, their product is positive too, not good for our inequality.

So, remaining points are #[-3,1]# and #5#.