Should we square it?

Here's my solution. First, the constraint of $x$ is $x>-3.0959...$ $\sqrt{x^2-x+1}=\frac{x^3+2x^2-3x+1}{x^2+2}$ $\Leftrightarrow (x+2)-\sqrt{x^2-x+1}+\frac{x^3+2x^2-3x+1}{x^2+2}-(x+2)=0$ $\Leftrightarrow \frac{x^2+4x+2-x^2+x-1}{x+2+\sqrt{x^2-x+1}}+\frac{x^3+2x^2-3x+1-x^3-2x^2-2x-4}{x^2+2}=0$ $\Leftrightarrow (5x+3)\bigg(\frac{1}{x+2+\sqrt{x^2-x+1}}-\frac{1}{x^2+2}\bigg)=0$ So we have the first root $x=\frac{-3}{5}$ . Now we consider $\frac{1}{x+2+\sqrt{x^2-x+1}}-\frac{1}{x^2+2}=0$ $\Rightarrow x^2+2=x+2+\sqrt{x^2-x+1}$ $\Leftrightarrow x^2-x=\sqrt{x^2-x+1}$ $\Leftrightarrow (x^2-x)^2=x^2-x+1$ $\Leftrightarrow (x^2-x)^2-(x^2-x)-1=0$ We get $x^2-x=\frac{1+\sqrt{5}}{2}$ and $x^2-x=\frac{1-\sqrt{5}}{2}$ , solving these equation we get $x=\frac{1\pm\sqrt{3+2\sqrt{5}}}{2}$ . Finally, the product is $\approx 0.97$