Автор - rehberhhre
2x^2 - x - 1 ≥ 0
2x^2 - x - 1 = 0
D = 1 + 4*2 = 9
x1 = (1 + 3)/4 = 1
x2 = (1 - 3)/4 = - 2/4 = - 1/2
(x + 1/2)(x - 1) ≥ 0
+ - +
-------------- ( - 1/2) ---------------------(1) -------------> x
x ∈ ( - ∞; - 1/2] ∪ [1; + ∞)