#-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]#