TeX Quellcode:

\begin{array}{rclll}\dfrac{3z+4}{27z^2+72z+48}+\dfrac{z-6}{6z} &=& \dfrac{1}{3z+4} \cr\cr \dfrac{3z+4}{3\left(9z^2+24z+16\right)}+\dfrac{z-6}{6z} &=& \dfrac{1}{3z+4} &\vert & -\dfrac{1}{3z+4} \cr\cr \dfrac{1}{3(3z+4)}+\dfrac{z-6}{6z}-\dfrac{1}{3z+4} &=& 0 \cr\cr \dfrac{2z}{6z(3z+4)}+\dfrac{(z-6)(3z+4)}{6z(3z+4)}-\dfrac{6z}{6z(3z+4)} &=& 0 \cr\cr \dfrac{2z}{6z(3z+4)}+\dfrac{3z^2+4z-18z-24}{6z(3z+4)}-\dfrac{6z}{6z(3z+4)} &=& 0 \cr\cr\dfrac{2z+3z^2-14z-24-6z}{6z(3z+4)} &=& 0 &\vert & \cdot 6z(3z+4)\cr\cr 3z^2-18z-24 &=& 0 &\vert& :3\cr z^2-6z-8 &=& 0 &\vert& \text{p-q-Formel} \cr z_{1,2} &=& 3\pm \sqrt{(-3)^2-(-8)} \cr z_{1,2} &=& 3 \pm \sqrt{17} \cr\cr z_{1,2} &=& 3+\sqrt{17} \approx 7{,}12 \;\in\;\mathbb{D} \cr z_{1,2} &=& 3-\sqrt{17} \approx -1{,}12 \;\in\;\mathbb{D} \cr\cr\mathbb{L} &=& \left\{3-\sqrt{17};3+\sqrt{17}\right\} \end{array}