(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`
From the graph we see that there are two solutions: `x=1/3,x=1`
This can be verified algebraically:
(2) Or you can rewrite the equation so it is equal to zero and graph to find the x-intercepts:
Graph `y=2|3x-2|-2` ;
Again we see that the solutions (here the x-intercepts) are `x=1/3,x=1`