TeX source:
\begin{array}{crclll}& \mathbb{D} &=& \mathbb{R}\setminus_{\{-1\}} \\\\& \dfrac{3}{\left(x+1\right)^2}\cdot\left(e^{\frac{4}{5}x+3}-3\right) &=& 0 \\\\\text{Faktor 1:} & \dfrac{3}{\left(x+1\right)^2} &=& 0 &\vert& \cdot \left(x+1\right)^2 \\& 3 &=& 0 \\\\\text{Faktor 2:} & e^{\frac{4}{5}x+3}-3 &=& 0 &\vert & +3 \\& e^{\frac{4}{5}x+3} &=& 3 &\vert & \ln() \\& \dfrac{4}{5}x+3 &=& \ln(3) &\vert & -3\\& \dfrac{4}{5}x &=& \ln(3) -3 &\vert & :\dfrac{4}{5}\\& x &=& \dfrac{5}{4}\left(\ln(3) -3\right) \approx -2{,}38\end{array}