(1) You can graph both `y=2|3x-2|-6` and `y=-4` and find the intersections.

The graph of y=-4 is a horizontal line.

The graph of `y=2|3x-2|-6` is a "V", opening up. Rewriting as:

`y=2|3(x-2/3)|-6=6|x-2/3|-6` we can see that the vertex is at `(2/3,-6)` and the slopes of the sides are `+-6`

The graphs:

From the graph we see that there are two solutions: `x=1/3,x=1`

This can be verified algebraically:

`2|3x-2|-6=-4`

`2|3x-2|=2`

`|3x-2|=1`

`3x-2=1==>3x=3==>x=1`

`3x-2=-1==>3x=1==>x=1/3`

(2) Or you can rewrite the equation so it is equal to zero and graph to find the x-intercepts:

`2|3x-2|-6=-4`

`2|3x-2|-2=0`

Graph `y=2|3x-2|-2` ;

Again we see that the solutions (here the x-intercepts) are `x=1/3,x=1`