[tex]x+\sqrt{x+2}-8=0\qquad(x\geq-2)\\\sqrt{x+2}=-x+8\qquad (x\leq8)\\x+2=(-x+8)^2\\x+2=x^2-16x+64\\x^2-17x+62=0\\\Delta=17^2-4\cdot1\cdot62=289-248=41\\\sqrt{\Delta}=\sqrt{41}\\x_1=\dfrac{-(-17)-\sqrt{41}}{2\cdot1}=\dfrac{17-\sqrt{41}}{2}\\x_2=\dfrac{-(-17)+\sqrt{41}}{2\cdot1}=\dfrac{17+\sqrt{41}}{2}\\\\x_2\not\leq8\\\\\boxed{x=\dfrac{17-\sqrt{41}}{2}}[/tex]
