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%%  Example 07-04-1 
%%  Copyright (C) 2017 Herbert Voss
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\documentclass[]{screxa}
\pagestyle{empty}
\setlength\textwidth{328.62788pt}
\setlength\parindent{0pt}
\usepackage[T1]{fontenc}
%StartShownPreambleCommands
\usepackage{amsmath}
%StopShownPreambleCommands
\begin{document}
\begin{equation}
\begin{array}{rcll}
y & = & x^{2}+bx+c\\
  & = & x^{2}+2\cdot{\displaystyle\frac{b}{2}x+c}\\
  & = & \underbrace{x^{2}+2\cdot\frac{b}{2}x+\left(\frac{b}{2}\right)^{2}}-{\displaystyle
 \left(\frac{b}{2}\right)^{2}+c}\\
  &  & \qquad\left(x+{\displaystyle \frac{b}{2}}\right)^{2}\\
  & = & \left(x+{\displaystyle \frac{b}{2}}\right)^{2}-\left({\displaystyle
     \frac{b}{2}}\right)^{2}+c & \left|+\left({\displaystyle
 \frac{b}{2}}\right)^{2}-c\right.\\
y+\left({\displaystyle \frac{b}{2}}\right)^{2}-c & = & \left(x+{\displaystyle
      \frac{b}{2}}\right)^{2} & \left|(\text{Scheitelpunktform})\right.\\
y-y_{S} & = & (x-x_{S})^{2}\\
S(x_{S};y_{S}) & \,\text{bzw.}\, & S\left(-{\displaystyle
    \frac{b}{2};\,\left({\displaystyle \frac{b}{2}}\right)^{2}-c}\right)
\end{array}
\end{equation}
\end{document}
