Beispiel Nr: 31
$ \text{Gegeben:} ax^{2}+bx+c=0 \\ \text{Gesucht:} \\ \text{Lösung der Gleichung} \\ \\ ax^{2}+bx+c=0 \\ \textbf{Gegeben:} \\ \frac{5}{9}x^2-5 =0 \\ \\ \textbf{Rechnung:} \\ \begin{array}{l|l|l|l} \begin{array}{l} \text{Umformen}\\ \hline \frac{5}{9}x^2-5 =0 \qquad /+5 \\ \frac{5}{9}x^2= 5 \qquad /:\frac{5}{9} \\ x^2=\displaystyle\frac{5}{\frac{5}{9}} \\ x=\pm\sqrt{9} \\ x_1=3 \qquad x_2=-3 \end{array}& \begin{array}{l} \text{a-b-c Formel}\\ \hline \\ \frac{5}{9}x^{2}+0x-5 =0 \\ x_{1/2}=\displaystyle\frac{-0 \pm\sqrt{0^{2}-4\cdot \frac{5}{9} \cdot \left(-5\right)}}{2\cdot\frac{5}{9}} \\ x_{1/2}=\displaystyle \frac{-0 \pm\sqrt{11\frac{1}{9}}}{1\frac{1}{9}} \\ x_{1/2}=\displaystyle \frac{0 \pm3\frac{1}{3}}{1\frac{1}{9}} \\ x_{1}=\displaystyle \frac{0 +3\frac{1}{3}}{1\frac{1}{9}} \qquad x_{2}=\displaystyle \frac{0 -3\frac{1}{3}}{1\frac{1}{9}} \\ x_{1}=3 \qquad x_{2}=-3 \end{array}& \begin{array}{l} \text{p-q Formel}\\ \hline \\ \frac{5}{9}x^{2}+0x-5 =0 \qquad /:\frac{5}{9} \\ x^{2}+0x-9 =0 \\ x_{1/2}=\displaystyle -\frac{0}{2}\pm\sqrt{\left(\frac{0}{2}\right)^2- \left(-9\right)} \\ x_{1/2}=\displaystyle 0\pm\sqrt{9} \\ x_{1/2}=\displaystyle 0\pm3 \\ x_{1}=3 \qquad x_{2}=-3 \end{array}\\ \end{array}$