- händische Berechnung mit der p-q-Formel
{ \displaystyle \begin{array}{l}f\left( x \right)\text{ }=\text{ }2{{x}^{2}}+\text{ }3x\text{ }-2\\{{x}_{0}}\,\,ist\,\,NST\,\Leftrightarrow \,f({{x}_{0}})\,=\,0\\\\0\,=\,2{{x}^{2}}+\text{ }3x\text{ }-2\left| :2 \right.\\0\,=\,\,{{x}^{2}}\,+\,\frac{3}{2}\,x\,-1\,\,\,\,\,Normalform\,\,\,\,\,0=\,{{x}^{2}}\,+\,px\,+\,q\\\\Einsetzen\,\,in\,\,die\,p-q-Formel\\{{x}_{1,\,2\,}}=\,-\frac{p}{2}\,\pm \,\sqrt{{{\left( \frac{p}{2} \right)}^{2}}\,-\,q\,\,}\\{{x}_{1,\,2\,}}=\,-\frac{\frac{3}{2}}{2}\,\pm \,\sqrt{{{\left( \frac{\frac{3}{2}}{2} \right)}^{2}}\,+\,1\,\,}\\{{x}_{1,\,2\,}}=\,-\frac{3}{4}\,\pm \,\sqrt{{{\left( \frac{3}{4} \right)}^{2}}\,+\,1\,\,}\\{{x}_{1,\,2\,}}=\,-\frac{3}{4}\,\pm \,\sqrt{\frac{9}{16}\,+\,\frac{16}{16}\,\,}\\{{x}_{1,\,2\,}}=\,-\frac{3}{4}\,\pm \,\sqrt{\frac{25}{16}}\\{{x}_{1,\,2\,}}=\,-\frac{3}{4}\,\pm \,\frac{5}{4}\\{{x}_{0,1\,}}=\,-\,2;\,\,\,\,{{x}_{0,2}}\,=\,0,5\end{array}}