about (scheduled) :: every now and then (not too often) i set myself to re-figure it out (the history has proven there exists a variance of what i come up with each time ...)
... so -- Def.-s , etc. ... ::
\[\begin{array}{lcl}
\left({x-a}\right)\left({x-b}\right)=0 &\ &\\
x^2-\left({a-b}\right)x+ab=0 &\ &\\
\begin{array}{l}
x^2+px+q=0\qquad \qquad \qquad \qquad \rightarrow\\
x^2+2px+p^2=px-q+p^2\ |×3 &\\
x^2-2px+p^2=-3px-q+p^2\ |+\ \uparrow &\\
\hline
4x^2+4px+4p^2=0-4q+4p^2\ |-3p^2\\
4x^2+4px+p^2=p^2-4q\ |÷4\\
\mathbf{x^2+2\frac p2x+{\left({\frac p2}\right)}^2={\left({\frac p2}\right)}^2-q}
\end{array} &\ &
\begin{array}{l}
\mathbf{x^2+2\frac p2x+{\left({\frac p2}\right)}^2={\left({\frac p2}\right)}^2-q}\\
{\left({x+\frac p2}\right)}^2={\left({\frac p2}\right)}^2-q\\
\boxed{x=-\frac p2±\sqrt{{\left({\frac p2}\right)}^2-q}}\\{}\\{}
\end{array}
\end{array}\]
... (it) came out double ((at) this time) -- the short and the long -- way to (the solution) F;T
// is likely ↑↑ why ↑↑ in many blogs folks do not get a thing what i say
// (as an old school programmer i always live-compact my code(read: text))
see also the inner properties of @ About the Quadratic Equation
[Eop]
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