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)#
