Помогитеее √х+1=8-2х

$\sqrt{x+1}=8-2x\\ =========================\\ \left \{ {{x+1 \geq 0} \atop {8-2x \geq 0}} \right. \\ \left \{ {{x \geq -1} \atop {-2x \geq -8}} \right. \\ \left \{ {{x \geq -1} \atop {x \leq 4}} \right. \Rightarrow x\in [-1;4] \\ =========================\\ ( \sqrt{x+1})^2=(8-2x)^2\\ x+1=64-32x+4x^2\\ 64-32x+4x^2-x-1=0\\ 4x^2-33x+63=0\\ D=b^2-4ac=(-33)^2-4\cdot4\cdot63=1089-1008=81=9^2\\ x_1= \frac{33+9}{8} = 5.25 \notin [-1;4] \\ x_2= \frac{33-9}{8} = 3\\$

Ответ: x = 3