2а-19=3а+7
а=-26
...............
(х - 3)^2 - 5|х - 3| + 4 ≤ 0
(x - 3)^2 = |x - 3|^2
|x - 3| = t ≥ 0
t^2 - 5t + 4 ≤ 0
D = 25 - 16 = 9 = 3^2
t12 = (5 +- 3)/2 = 1 4
(t - 1)(t - 4) ≤ 0
++++++++[1] ------------- [4] +++++++++++
1 ≤ t ≤ 4
1. |x - 3| ≥ 1
x - 3 ≥ 1 x ≥ 4
x - 3 ≤ -1 x ≤ 2
x ∈ (-∞,2] U [4, +∞)
2. |x - 3| ≤ 4
x - 3 ≤ 4 x ≤ 7
x - 3 ≥ -4 x ≥ -1
x ∈ [-1, 7]
пересекаем с первым вариантом
ответ х ∈ [-1, 2] U [4,7]