You can solve such problems by splitting into cases where the sign of the enclosed expressions changes, but on this occasion I will square both sides instead to get:

#4v^2 = 256+96v+9v^2#

Then subtract #4v^2# from both sides to get:

#5v^2+96v+256 = 0#

Next use a modified AC Method to find factors:

#A=5#, #B=96#, #C=256#

Look for a factorization of #AC=5*256# into a pair of factors whose sum is #B=96#.

With a little reasoning, I quickly found the pair #B1=16#, #B2=80#

Then for each combination of #A, B1# and #A, B2# divide by their HCF (highest common factor) to give a pair of coefficients of a factor:

#(A, B1) = (5, 16)# HCF #1# #-> (5, 16) -> (5v+16)#

#(A, B2) = (5, 80)# HCF #5# #-> (1, 16) -> (v+16)#

So #5v^2+96v+256 = (5v+16)(v+16)#

This has zeros for #v = -16# and #v = -16/5#

Check:

#v=-16# :

#abs(2v) = 32 = abs(-16+48) = abs(-16-3v)#

#v = -16/5# :

#abs(2v) = 32/5 = abs(-16+16-32/5)#

#= abs(-16+(80-32)/5) = abs(-16+48/5)#

#= abs(-16+(3*16)/5) = abs(-16-3v)#