First, expand the terms within parenthesis:

#(color(red)(-2) xx x) - (color(red)(-2) xx 4) <= 3x#

#-2x + 8 <= 3x#

Next, add #color(red)(2x)# to each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#color(red)(2x) - 2x + 8 <= color(red)(2x) + 3x#

#0 + 8 <= (2 + 3)x#

#8 <= 5x#

Now, divide each side of the inequality by #color(red)(5)# to solve for #x# while keeping the inequality balanced:

#8/color(red)(5) <= (5x)/color(red)(5)#

#8/5 <= (color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5))#

#8/5 <= x#

To put the solution in terms of #x# we can reverse or "flip" the inequality:

#x >= 8/5#