x + 2\\x + 2 \geq 0; x\geq -2\\x + 8 > x^2 + 4x +4\\x^2 + 3x - 4 < 0\\D = 9 + 16 = 25\\x_1 = \frac{-3 + 5}2 = 1; x_2 = \frac{-3-5}2 = -4" alt="\sqrt{x+8} > x + 2\\x + 2 \geq 0; x\geq -2\\x + 8 > x^2 + 4x +4\\x^2 + 3x - 4 < 0\\D = 9 + 16 = 25\\x_1 = \frac{-3 + 5}2 = 1; x_2 = \frac{-3-5}2 = -4" align="absmiddle" class="latex-formula">
⇒ x ∈ [-2; 1)