【解答】
\colorbox{mistyrose}{$3$}x^2\colorbox{lightcyan}{$-7$}x\colorbox{lightgreen}{$+1$}=0\def\ka{3} \def\kb{-7} \def\kc{1} \def\kna{6} \def\kmb{7} \def\kbz{49} \def\kyac{-12} \def\kr{37} \begin{align*} x &= \dfrac{-\colorbox{lightcyan}{$(\kb)$}\pm\sqrt{\colorbox{lightcyan}{$(\kb)$}^2-4 \cdot \colorbox{mistyrose}{$\ka$}\cdot\colorbox{lightgreen}{$\kc$}}}{2 \cdot \colorbox{mistyrose}{$\ka$}}\\\\ &= \dfrac{\kmb \pm \sqrt{\kbz \kyac}}{\kna}\\ \\ &= \dfrac{\kmb \pm \sqrt{\kr}}{\kna}\\ \\ \end{align*}
【解答】
\colorbox{mistyrose}{$1$}x^2\colorbox{lightcyan}{$+7$}x\colorbox{lightgreen}{$+4$}=0\def\ka{1} \def\kb{7} \def\kc{4} \def\kna{2} \def\kmb{-7} \def\kbz{49} \def\kyac{-16} \def\kr{33} \begin{align*} x &= \dfrac{-\colorbox{lightcyan}{$\kb$}\pm\sqrt{\colorbox{lightcyan}{$\kb$}^2-4 \cdot \colorbox{mistyrose}{$\ka$}\cdot\colorbox{lightgreen}{$\kc$}}}{2 \cdot \colorbox{mistyrose}{$\ka$}}\\\\ &= \dfrac{\kmb \pm \sqrt{\kbz \kyac}}{\kna}\\ \\ &= \dfrac{\kmb \pm \sqrt{\kr}}{\kna}\\ \\ \end{align*}
【解答】
\colorbox{mistyrose}{$3$}x^2\colorbox{lightcyan}{$+5$}x\colorbox{lightgreen}{$-1$}=0\def\ka{3} \def\kb{5} \def\kc{(-1)} \def\kna{6} \def\kmb{-5} \def\kbz{25} \def\kyac{+12} \def\kr{37} \begin{align*} x &= \dfrac{-\colorbox{lightcyan}{$\kb$}\pm\sqrt{\colorbox{lightcyan}{$\kb$}^2-4 \cdot \colorbox{mistyrose}{$\ka$}\cdot\colorbox{lightgreen}{$\kc$}}}{2 \cdot \colorbox{mistyrose}{$\ka$}}\\\\ &= \dfrac{\kmb \pm \sqrt{\kbz \kyac}}{\kna}\\ \\ &= \dfrac{\kmb \pm \sqrt{\kr}}{\kna}\\ \\ \end{align*}
【解答】
\colorbox{mistyrose}{$3$}x^2\colorbox{lightcyan}{$-8$}x\colorbox{lightgreen}{$-3$}=0\def\ka{3} \def\kb{(-8)} \def\kc{(-3)} \def\kna{6} \def\kmb{8} \def\kbz{64} \def\kyac{+36} \def\kr{100} \begin{align*} x &= \dfrac{-\colorbox{lightcyan}{$\kb$}\pm\sqrt{\colorbox{lightcyan}{$\kb$}^2-4 \cdot \colorbox{mistyrose}{$\ka$}\cdot\colorbox{lightgreen}{$\kc$}}}{2 \cdot \colorbox{mistyrose}{$\ka$}}\\\\ &= \dfrac{\kmb \pm \sqrt{\kbz \kyac}}{\kna}\\ \\ &= \dfrac{\kmb \pm \sqrt{\kr}}{\kna}\\ \\ &= \dfrac{\kmb \pm 10}{\kna}\\ \\ &= \dfrac{\kmb + 10}{\kna},\ \dfrac{\kmb - 10}{\kna}\\ \\ &= \dfrac{18}{\kna},\ \dfrac{-2}{\kna}\\ \\ &= 3,\ -\dfrac{1}{3} \end{align*}
【解答】
\colorbox{mistyrose}{$9$}x^2\colorbox{lightcyan}{$-12$}x\colorbox{lightgreen}{$+4$}=0\def\ka{9} \def\kb{(-12)} \def\kc{4} \def\kna{18} \def\kmb{12} \def\kbz{144} \def\kyac{-144} \def\kr{0} \begin{align*} x &= \dfrac{-\colorbox{lightcyan}{$\kb$}\pm\sqrt{\colorbox{lightcyan}{$\kb$}^2-4 \cdot \colorbox{mistyrose}{$\ka$}\cdot\colorbox{lightgreen}{$\kc$}}}{2 \cdot \colorbox{mistyrose}{$\ka$}}\\\\ &= \dfrac{\kmb \pm \sqrt{\kbz \kyac}}{\kna}\\ \\ &= \dfrac{\kmb \pm \sqrt{\kr}}{\kna}\\ \\ &= \dfrac{\kmb \pm 0}{\kna}\\ \\ &= \dfrac{\kmb}{\kna}\\ \\ &= \dfrac{2}{3} \end{align*}