#-2x-4<=(x+2)^2<=-2x-1# Adding #2x# to each inequality term
#-4 le (x+2)^2+2x le -1# so the solution set is for all #x# that simultaneously
1)
#-4 le (x+2)^2+2x->0 le (x+2)^2+2x+4#
and
2)
# (x+2)^2+2x le -1-> (x+2)^2+2x+1 le 0#
Solving 1)
the roots for
#x^2+6x+8 = 0#
are
#-4,-2#
so for
#x le -4# and # x ge -2# we have #x^2+6x+8 ge 0#
Solving 2)
the roots for
#(x+2)^2+2x+1=0#
are
#-5,-1#
so for
#-5 le x le-1# we have #(x+2)^2+2x+1 le 0#
and the intersection set is
#-5 le x le -4# and #-2 le x le -1#
so the solution set is
#x in [-5,-4] uu [-2,-1]#
