√(5x² + 4) ≤ 7x + 10
(√(5x² + 4))² ≤ (7x + 10)²
5x² + 4 ≤ 49x² + 140x + 100
5x² - 49x² - 140x + 4 - 100 ≤ 0
-44x² - 140x - 96 ≤ 0
-44x² - 140x - 96 = 0 | : (-4)
11x² + 35x + 24 = 0
D = b² - 4ac = 35² - 4 × 11 × 24 = 1225 - 1056 = 169
x₁,₂ = (-b ± √D) ÷ 2a
x₁ = (-35 + 13) ÷ 2 × 11 = -48 ÷ 22 = -24/11 = -2, 2/11 (минус 2 целых 2 одинадцатых)
x₂ = (-35 - 13) ÷ 2 × 11 = -22 ÷ 22 = -1
x " alt="--(minus)--(-2\frac{2}{11})--(plus)--(-1)--(minus)-->x " align="absmiddle" class="latex-formula">
x ≤ -2, 2/11
x ≥ -1
Ответ: x ∈ (-∞; -2, 2/11] U [-1; +∞)