When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

$$ A= \begin{bmatrix} 1 & 2 &3 \\ 4 &5 & 6\\ 7 & 8 &9 \end{bmatrix} $$

$$ \begin{vmatrix} A \end{vmatrix}= \begin{vmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}=0 $$

$$ \begin{vmatrix} A-\lambda I \end{vmatrix}= \begin{vmatrix} 1-\lambda & 2 & 3\\ 4 &5-\lambda &6 \\ 7 &8 & 9-\lambda \end{vmatrix} =0 $$