If anyone could help me see how $$ A(\xi,\eta)=a\,\xi_x^2+2\,b\,\xi_x\,\xi_y+c\,\xi_y^2=0 $$ is turned into $$ \frac{1}{a} \left[a\,\xi_x+\left(b-\sqrt{b^2-ac}\right)\xi_y\right] \left[a\,\xi_x+\left(b+\sqrt{b^2-ac}\right)\xi_y\right]=0. $$
Thanks in advance!
$$ \begin{align} &\frac{1}{a} \left[a\,\xi_x+\left(b-\sqrt{b^2-ac}\right)\xi_y\right] \left[a\,\xi_x+\left(b+\sqrt{b^2-ac}\right)\xi_y\right]=\\ =&\frac{1}{a}\left\{a^2\,\xi_x^2+ a\left(b\color{red}{+\sqrt{b^2-ac}}\right)\xi_x\xi_y+ a\left(b\color{red}{-\sqrt{b^2-ac}}\right)\xi_x\xi_y+ \left[b^2-\left(\sqrt{b^2-ac}\right)^2\right]\xi_y^2\right\}=\\ =&\frac{1}{a} \left\{a^2\,\xi_x^2+2\,a\,b\,\xi_x\,\xi_y+ \left[\color{blue}{b^2}-\left(\color{blue}{b^2}-ac\right)\right]\xi_y^2\right\}=\\ =&\frac{1}{a} \left\{a^2\,\xi_x^2+2\,a\,b\,\xi_x\,\xi_y+a\,c\,\xi_y^2\right\}=\\ =&a\,\xi_x^2+2\,b\,\xi_x\,\xi_y+c\,\xi_y^2. \end{align} $$ Or, if you prefer, you can read this bottom to top.