Пусть 2x^2 + 3x = t
√(t + 9) = 33 - t
t + 9 = t^2 - 66t + 1089
t^2 - 67t + 1080 = 0
D = 169 = 13^2
t₁ = (67 + 13)/2 = 40 ∉ t ≤ 33
t₂ = (67 - 13)/2 = 27 ∈ t ≤ 33
2x^2 + 3x - 27 = 0
D = 225 = 15^2
x₁= ( - 3 + 15)/4 = 3 ∈ - 4,88≤ x ≤ 3.38
x₂ = ( - 3 - 15)/4 = - 4,5 ∈ - 4,88 ≤ x ≤ 3.38
Ответ
- 4,5 ; 3