(scheduled) derivation of the solution of the quadratic equation

 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

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[Eop]