(2x-6)² ≥ (6x-2)²
|2x-6| ≥ |6x - 2|
2x - 6 = 0
x = 3
6x - 2 = 0
x = 1/3
1) x ≤ 1/3
-2x + 6 ≥ -6x + 2
4x ≥ -4
x ≥ -1 => x∈[-1; 1/3]
2) 1/3 < x < 3
-2x + 6 ≥ 6x - 2
8x ≤ 8
x ≤ 1 => x∈(1/3; 1]
3) x ≥ 3
2x - 6 ≥ 6x - 2
4x ≤ -4
x ≤ -1 - не подходит
Ответ: x∈[-1; 1]