X^2 + 2|x| - 3 <=0 решить неравенство

1) x ≥ 0

x² + 2x - 3 ≤ 0

D = 4 + 12 = 16

x1 = (-2-4) / 2 = - 3

x2 = (-2+4) / 2 = 1

x∈[0; 1]

2) x < 0

x² - 2x - 3 = 0

D = 4 + 12 = 16

x1 = (2-4) / 2 = - 1

x2 = (2+4) / 2 = 3

x∈[-1; 0)

Общий ответ: x∈[-1; 1]