We have to solve: 4*|2x - 6| + 8 = 12

4*|2x - 6| + 8 = 12

=> 4*(2x - 6) + 8 = 12 and 4(6 - 2x) + 8 = 12

=> 8x - 24 + 8 = 12 and 24 - 8x + 8 = 12

=> 8x = 12 + 24 - 8 and -8x = 12 - 8 - 24

=> 8x = 28 and -8x = -20

=> x = 28/8 and x = 20/8

=> x = 3.5 and x = 2.5

**The required values of x are x = 3.5 and x = 2.5**

We'll solve the absolute value equation as it follows:

4 | 2x-6 | + 8 = 12

We'll first subtract 8 both sides in order to isolate the absolute value to the left side:

4 | 2x-6 | = 12 - 8

4 | 2x-6 | = 4

We'll divide by 4:

| 2x-6 | = 1

We'll get 2 cases to solve:

1) We'll impose the constraint of absolute value:

2x-6>=0

2x>=6

x>=3

Now, we'll solve the equation:

2x-6 = 1

We'll add 6 both sides:

2x = 7

x = 7/2 = 3.5 > 3

The value of x belongs to the interval of admissible values:

[7/2 , +inf.)

2) 2x-6<0

2x<6

x<3

Now, we'll solve the equation:

2x-6 = -1

We'll subtract 6 both sides:

2x = 5

We'll divide by 2:

x = 5/2

Since the value of x belongs to the interval of admissible values, x = 2/5 is also a root of the given equation.

**Since both values of x respect the constraints of the modulus, they represent solutions of the equation: {5/2 ; 7/2}.**