【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*}