Algebra-Gleichungen-Quadratische Gleichung

$ ax^{2}+bx+c=0 $
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Beispiel Nr: 03
$\begin{array}{l} \text{Gegeben:} ax^{2}+bx+c=0 \\ \text{Gesucht:} \\ \text{Lösung der Gleichung} \\ \\ ax^{2}+bx+c=0 \\ \textbf{Gegeben:} \\ -\frac{2}{3}x^2+\frac{1}{6} =0 \\ \\ \textbf{Rechnung:} \\ \begin{array}{l|l|l|l} \begin{array}{l} \text{Umformen}\\ \hline -\frac{2}{3}x^2+\frac{1}{6} =0 \qquad /-\frac{1}{6} \\ -\frac{2}{3}x^2= -\frac{1}{6} \qquad /:\left(-\frac{2}{3}\right) \\ x^2=\displaystyle\frac{-\frac{1}{6}}{-\frac{2}{3}} \\ x=\pm\sqrt{\frac{1}{4}} \\ x_1=\frac{1}{2} \qquad x_2=-\frac{1}{2} \end{array}& \begin{array}{l} \text{a-b-c Formel}\\ \hline \\ -\frac{2}{3}x^{2}+0x+\frac{1}{6} =0 \\ x_{1/2}=\displaystyle\frac{-0 \pm\sqrt{0^{2}-4\cdot \left(-\frac{2}{3}\right) \cdot \frac{1}{6}}}{2\cdot\left(-\frac{2}{3}\right)} \\ x_{1/2}=\displaystyle \frac{-0 \pm\sqrt{\frac{4}{9}}}{-1\frac{1}{3}} \\ x_{1/2}=\displaystyle \frac{0 \pm\frac{2}{3}}{-1\frac{1}{3}} \\ x_{1}=\displaystyle \frac{0 +\frac{2}{3}}{-1\frac{1}{3}} \qquad x_{2}=\displaystyle \frac{0 -\frac{2}{3}}{-1\frac{1}{3}} \\ x_{1}=-\frac{1}{2} \qquad x_{2}=\frac{1}{2} \end{array}& \begin{array}{l} \text{p-q Formel}\\ \hline \\ -\frac{2}{3}x^{2}+0x+\frac{1}{6} =0 \qquad /:-\frac{2}{3} \\ x^{2}+0x-\frac{1}{4} =0 \\ x_{1/2}=\displaystyle -\frac{0}{2}\pm\sqrt{\left(\frac{0}{2}\right)^2- \left(-\frac{1}{4}\right)} \\ x_{1/2}=\displaystyle 0\pm\sqrt{\frac{1}{4}} \\ x_{1/2}=\displaystyle 0\pm\frac{1}{2} \\ x_{1}=\frac{1}{2} \qquad x_{2}=-\frac{1}{2} \end{array}\\ \end{array} \end{array}$