How do you solve #abs(20+1/2x)>6#?

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

#-6 > 20 + 1/2x > 6#

#-6 - color(red)(20) > 20 - color(red)(20) + 1/2x > 6 - color(red)(20)#

#-26 > 0 + 1/2x > -14#

#color(red)(2) xx -26 > color(red)(2) xx 1/2x > color(red)(2) xx -14#

#-52 > color(red)(2)/2x > -28#

#-52 > 1x > -28#

#-52 > x > -28#

Or

#x < -52#; #x > -28#

Or, in interval notation:

#(-oo, -52)#; #(-28, +oo)#

One way to approach the evaluation of an absolute value inequality like #abs(20+1/2x) > 6# is to first temporarilly replace #(20+1/2x)# with some other variable.
For example if we let #z=(20+1/2x)# then we have the simpler inequality:
#color(white)("XXX")abs(z) > 6#
Presented on the number line this would look like:

That is
#color(white)("XXXxxxxxx")z < -6color(white)("xxxx")"or"color(white)("xxxxxxxxx") z > +6#
then since we know #z=20+1/2x#
these become
#color(white)("XXX")20+1/2x < -6color(white)("xxxx")"or"color(white)("xxxx") 20+1/2x > +6#

#color(white)("XXX20+")1/2x < -26color(white)("xxxx")"or"color(white)("xxxx20+") 1/2x > -14#

#color(white)("XXX"20+2)x < -52color(white)("xxxx"20+2)"or"color(white)("xxxx") x > -28#