Incidentally you can type absolute value bars || using the
key just above the ENTER or RETURN key.  It is the shifted \. 

|3x+2|-|x-3| = 5

There are 4 cases to consider, which are all 4 ways to
place signs in front on the expressions between the
absolute value bars:

+(3x+2)+(x-3) = 5, +(3x+2)-(x-3) = 5, -(3x+2)+(x-3) = 5, -(3x+2)-(x-3) = 5
     3x+2+x-3 = 5,      3x+2-x+3 = 5,     -3x-2+x-3 = 5,     -3x-2-x+3 = 5
         4x-1 = 5,          2x+5 = 5,         -2x-5 = 5,         -4x+1 = 5
           4x = 6,            3x = 0,           -2x = 10,          -4x = 4
            x = 6/4,           x = 0              x = -5             x = -1
            x = 3/2

We must check for extraneous solutions by substituting into the
original equation:

Checking x = 3/2

       |3x+2|-|x-3| = 5
 |3(3/2)+2|-|3/2-3| = 5
|9/2+4/2|-|3/2-6/2| = 5
      |13/2|-|-3/2| = 5
       (13/2)-(3/2) = 5
               10/2 = 5
                  5 = 5

So 3/2 is a solution.

Checking x = 0

       |3x+2|-|x-3| = 5
 |3(0)+2|-|0-3| = 5
     |0+2|-|-3| = 5
          |2|-3 = 5
            2-3 = 5
             -1 = 5

So 0 is NOT a solution.
            
Checking x = -5

       |3x+2|-|x-3| = 5
   |3(-5)+2|-|-5-3| = 5
       |-15+2|-|-8| = 5
            |-13|-8 = 5
               13-8 = 5
                  5 = 5

So -5 is a solution.

Checking x = -1

       |3x+2|-|x-3| = 5
   |3(-1)+2|-|-1-3| = 5
        |-3+2|-|-4| = 5
             |-1|-4 = 5
                1-4 = 5
                 -3 = 5

So -1 is NOT a solution

The only solutions are 3/2 and -5

Edwin