The answer: #{-14/5,2}#

Absolute value operation always return non-negative.

So,

#if (5y+2)/2>=0 ->abs((5y+2)/2) = (5y+2)/2#

#if (5y+2)/2<0 ->abs((5y+2)/2) = (-1) * ((5y+2)/2)#

Since we don't know the value of #y#, we need to check all these possibilities.

**First**

Assume that #(5y+2)/2 >= 0#

We need to find the range of #y#

#2 * ((5y+2)/2) >= 2 * 0#

#5y +2 >= 0#

#A: y>=-2/5#

We need to remember this.

Now, try to solve the given equation.

#(5y+2)/2 = 6#

#5y+2=12#

#5y=10#

#y = 2#

Remember expression #A#, #y=2# satisfies the inequality in expression #A#. So #2# is in the solution set.

**Second**

Assume that #(5y+2)/2<0#

#B: y < -2/5#

Now try to solve the given equation again, but this time the absolute value operation will return #(-1)# times of the input since the input is assumed negative.

#(-1) * (5y+2)/2 = 6#

#(5y+2)/2 = -6#

#5y+2 = -12#

#5y = -14#

#y = -14/5#

#y=-14/5# satisfies the inequality in the expression #B# so #-14/5# is also in our solution set.

Result: #{-14/5,2}#