Помогите, пожалуйста, один маленький номерок решииить:3

Решите задачу:

$x^2+2x+\sqrt{x^2+2x+8}=12\\t=x^2+2x\\\\t+ \sqrt{t+8}=12 \\ \sqrt{t+8}=12-t\\\\ \left \{ {{12-t \geq 0} \atop {t+8=(12-t)^2}} \right. \\\\ \left \{ {{t \leq 12} \atop {t+8=144-24t+t^2}} \right. \\\\ \left \{ {{t \leq 12} \atop {t^2-25t+136=0}} \right. \\\\D=81\\t_{1}=17, t_{2}=8$

$x^2+2x=17\\x^2+2x-17=0\\D=2^2-4*1*(-17)=72\\ x_{1}= \frac{-2+ \sqrt{72} }{2}= \frac{-2+6 \sqrt{2} }{2}= -1+3 \sqrt{2}\\x_{2}=-1-3 \sqrt{2} \\\\x^2+2x=8\\x^2+2x-8=0\\x_{3}=2\\x_{4}=-4\\\\\\-1б3 \sqrt{2};2;-4$