Autumn tournament

Let $a=\sqrt{x^2+2x+2}>0$ and $b=\sqrt{x^2+1}>0$. Therefore $$x=\frac{(x^2+2x+2)-(x^2+1)-1}{2}=\frac{a^2-b^2-1}{2}$$ $$x+1=\frac{a^2-b^2+1}{2}.$$ The equation is equivalent to: $$\frac{a^2-b^2+1}{2}\cdot a+\frac{a^2-b^2-1}{2}\cdot b=0$$ $$(a^2-b^2)a+a+(a^2-b^2)b-b=0$$ $$(a-b)((a+b{)}^2+1)=0.$$ Hence, $a=b$ and $x=-\frac{1}{2}$. $\square$