$\lefteqn{\overbrace{\phantom{1+2+3}}}1+\underbrace{2+3+4}$ $\makebox[0pt][l]{\(\overbrace{\phantom{1+2+3}}^{\textit{первые}}\)}1+\underbrace{2+3+4}_{\textit{вторые}}$ $\makebox[0pt][l]{$\sqrt{4}$}% \makebox[0pt][l]{{\color{blue}$\sqrt{\vphantom{4}\hphantom{4}}$}}% \hphantom{{\sqrt{4}}}% = \makebox[0pt][l]{$\sqrt{\dfrac{8}{2}}$}% \makebox[0pt][l]{{\color{green}$\sqrt{\vphantom{\dfrac{8}{2}}\hphantom{\dfrac{8}{2}}}$}}% \hphantom{{\sqrt{\dfrac{8}{2}}}}% $ $$x^2+px+q=x^2+2\cdot\left(\frac{p}2}\right)\cdot x+q= \makebox[0pt][l]{\color{magenta}\underbrace{\hphantom{x^2+2\left(\frac{p}2\right)+ \left(\frac{p}2\right)^2}\vphantom{x^2+2\left(\frac{p}2\right) x+\left(\frac{p}2\right)^2}}_{\makebox[0pt][c]{полный квадрат получился!}}} x^2+2\left(\frac{p}2}\right) x+\overbrace{\left(\frac{p}2\right)^2-\left(\frac{p}2\right)^2}^\substack{\text{\color{blue}чисто}\\ \text{\color{blue}нолик прибавили!}}} + q =\left(x+\frac{p}2}\right)^2-\left(\frac{p}2}\right)^2+q.$$