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next (failed?) try to calculate the quadratic
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@ -62,7 +62,7 @@ $b, c, d \in \mdr$ be a function.
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Then you could solve the following problem for $x$:
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\begin{align}
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0 &\stackrel{!}{=} \left ((d_{P,f}(x))^2 \right )'
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0 &\stackrel{!}{=} \left ((d_{P,f}(x))^2 \right )'\\
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&=-2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
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&= 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
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&= f(x) \cdot f'(x) - y_p f'(x) + x - x_p\\
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