Beispiel Nr: 39
$ \text{Gegeben:} ax^{2}+bx+c=0 \\ \text{Gesucht:} \\ \text{Lösung der Gleichung} \\ \\ ax^{2}+bx+c=0 \\ \textbf{Gegeben:} \\ -\frac{7}{9}x^2+4\frac{2}{3}x =0 \\ \\ \textbf{Rechnung:} \\ \begin{array}{l|l|l|l} \begin{array}{l} \text{x-Ausklammern}\\ \hline -\frac{7}{9}x^{2}+4\frac{2}{3}x =0 \\ x(-\frac{7}{9}x +4\frac{2}{3})=0 \\ \\ -\frac{7}{9} x+4\frac{2}{3} =0 \qquad /-4\frac{2}{3} \\ -\frac{7}{9} x= -4\frac{2}{3} \qquad /:\left(-\frac{7}{9}\right) \\ x=\displaystyle\frac{-4\frac{2}{3}}{-\frac{7}{9}}\\ x_1=0\\ x_2=6 \end{array}& \begin{array}{l} \text{a-b-c Formel}\\ \hline \\ -\frac{7}{9}x^{2}+4\frac{2}{3}x+0 =0 \\ x_{1/2}=\displaystyle\frac{-4\frac{2}{3} \pm\sqrt{\left(4\frac{2}{3}\right)^{2}-4\cdot \left(-\frac{7}{9}\right) \cdot 0}}{2\cdot\left(-\frac{7}{9}\right)} \\ x_{1/2}=\displaystyle \frac{-4\frac{2}{3} \pm\sqrt{21\frac{7}{9}}}{-1\frac{5}{9}} \\ x_{1/2}=\displaystyle \frac{-4\frac{2}{3} \pm4\frac{2}{3}}{-1\frac{5}{9}} \\ x_{1}=\displaystyle \frac{-4\frac{2}{3} +4\frac{2}{3}}{-1\frac{5}{9}} \qquad x_{2}=\displaystyle \frac{-4\frac{2}{3} -4\frac{2}{3}}{-1\frac{5}{9}} \\ x_{1}=0 \qquad x_{2}=6 \end{array}& \begin{array}{l} \text{p-q Formel}\\ \hline \\ -\frac{7}{9}x^{2}+4\frac{2}{3}x+0 =0 \qquad /:-\frac{7}{9} \\ x^{2}-6x+0 =0 \\ x_{1/2}=\displaystyle -\frac{-6}{2}\pm\sqrt{\left(\frac{\left(-6\right)}{2}\right)^2- 0} \\ x_{1/2}=\displaystyle 3\pm\sqrt{9} \\ x_{1/2}=\displaystyle 3\pm3 \\ x_{1}=6 \qquad x_{2}=0 \end{array}\\ \end{array}$