f(x) = abs(x+2)+abs(2x-4)-abs(x-3)=0
Making abs(x+2) = y = abs(x-3)-abs(2x-4)=0
we have x = pm y-2.
I) Choosing x = y-2
y = abs(y-5)-abs(2y-8)
y-5 = abs(y-5)-abs(2y-8)-5
1 = pm1-(abs(2y-8)+5)/(y-5) for y ne 5
Here we have two possibilities:
I-1)
0 = abs(2y-8)+5 having two outcomes
I-1-a)
2y-8=-5->y=3/2
I-1-b)
2y-8=5->y=13/2
I-2)
2=-( abs(2y-8)+5)/(y-5)
2(y-5)=-abs(2y-8)-5
2y-10=-abs(2y-8)-5
2y-8=-abs(2y-8)-3
1=pm 1-3/(2y-8)
with the only posibility
2 =- 3/(2y-8)->y = 13/4
II) Choosing x = -(y+2)
-y+abs(2y+8)-abs(y+5)=0
-y+2abs(y+4)-abs(y+5)=0
-y-4+2abs(y+4)-abs(y+5)+4=0
-(y+4)+2abs(y+4)-abs(y+5)+4=0
1 pm2 +(abs[y+5]-4) /(y+4) = 0 with two outcomes
II-1)
3 +(abs[y+5]-4) /(y+4) = 0
3(y+4)+abs(y+5)-4=0
3(y+5)+abs(y+5)-4-3=0
3pm1 -7/(y+5)=0 with two outcomes
II-1-a)
4(y+5)=7->y=-13/4
II-1-b)
2(y+5)=7->y=-3/2
II-2)
-1 +(abs[y+5]-4) /(y+4) = 0
-(y+4)+abs(y+5)-4=0
-(y+5)+abs(y+5)-4+1=0
1pm1+3/(y+5)=0
with the only outcome
-2(y+5)+3=0->y=-7/2
Collecting the results obtained for x = y-2
y = {3/2,13/2,13/4}->x={-1/2,9/2,5/4}
and for x = -(y+2)
y = {-13/4,-3/2,-7/2}->x={5/4,-1/2,3/2}
Substituting in f(x) none of them is solution.