geg.: $\mathrm{0,5}\cdot x^2+x-4=0\text{mit}x\in \mathbb{R}0\{,\}5\; \backslash cdot\; x^2+x-4=0\backslash qquad\; \backslash text\{mit\}\backslash :\; x\backslash in\; \backslash mathbb\{R\}$

ges.: Lösungsmenge für $xx$

Ansatz und Rechnung:

abc-Formel $x_{\mathrm{1,2}}=\frac{-b\pm \sqrt{{\mathbf{b}}^{\mathbf{2}}-\mathbf{4}\cdot \mathbf{a}\cdot \mathbf{c}}}{2\cdot a}\Rightarrow x_{\mathrm{1,2}}=\frac{-b\pm \sqrt{\mathbf{D}}}{2\cdot a}x\_\{1\{,\}2\}=\backslash frac\{-b\backslash pm\backslash sqrt\{\backslash bold\; \{b^2-4\backslash cdot\; a\; \backslash cdot\; c\}\}\}\{2\backslash cdot\; a\}\; \backslash Rightarrow\; x\_\{1\{,\}2\}=\backslash frac\{-b\backslash pm\backslash sqrt\{\backslash bold\; \{D\}\}\}\{2\backslash cdot\; a\}$

auf quadratische Gleichung $\mathrm{0,5}\cdot x^2+x-4=00\{,\}5\; \backslash cdot\; x^2+x-4=0$ anwenden:

mit $a=\mathrm{0,5}a=0\{,\}5$; $b=1\backslash quad\; b=1$; $c=-4\backslash quad\; c=-4$

$\Rightarrow D=b^2-4\cdot a\cdot c=1^2-4\cdot \mathrm{0,5}\cdot (-4)=1+8=9\backslash Rightarrow\; D=b^2-4\backslash cdot\; a\; \backslash cdot\; c=1^2-4\backslash cdot\; 0\{,\}5\backslash cdot\; (-4)=1+8=9$

$\Rightarrow D>0\leftrightarrow \backslash Rightarrow\; D>0\backslash leftrightarrow$ $22$ Lösungen

$\Rightarrow x_{\mathrm{1,2}}=\frac{-b\pm \sqrt{D}}{2\cdot a}=\frac{-1\pm \sqrt{9}}{1}=\frac{-1\pm 3}{1}=-1\pm 3\backslash Rightarrow\; x\_\{1\{,\}2\}=\backslash frac\{-b\backslash pm\backslash sqrt\{D\}\}\{2\backslash cdot\; a\}=\backslash frac\{-1\backslash pm\backslash sqrt\{9\}\}\{1\}=\backslash frac\{-1\backslash pm\; 3\}\{1\}=-1\; \backslash pm\; 3$

$\Rightarrow x_1=-1-3=-4\backslash Rightarrow\; x\_\{1\}=-1-3=-4$

$\Rightarrow x_2=-1+3=2\backslash Rightarrow\; x\_\{2\}=-1+3=2$

$\mathbb{L}=\{-\mathbf{4};\mathbf{2}\}\backslash bold\; \{\backslash mathbb\{L\}=\backslash left\backslash \{-4;\; 2\backslash right\backslash \}\}$