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

\begin{array}{clr} a+d & c & 9 \\ 5 & x^2+y & 3 \\ 8 & 7 & z_1 \\ \end{array}

$$\left\{\begin{array}{l} \dot{y}_{1}=y_{1}+2 y_{2} \\ \dot{y}_{2}=4 y_{1}+3 y_{2} \end{array}\right.$$