二次方程式の解の公式

$$\begin{eqnarray} ax^2+bx+c&=&0 \\x^2+\frac{b}{a}x+\frac{c}{a}&=&0 \\x^2+\frac{b}{a}x+\color{red}{\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2}\color{black}{}+\frac{c}{a}&=&0 \;\cdots\;\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2=0 \\x^2+\color{red}{2}\color{black}{}\frac{b}{\color{red}{2}\color{black}{}a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}\frac{4a}{4a}&=&0 \\\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2}&=&0 \;\cdots\;(x+a)^2=x^2+2ax+a^2 \\\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}&=&0 \\\left(x+\frac{b}{2a}\right)^2&=&\frac{b^2-4ac}{4a^2} \\x+\frac{b}{2a}&=&\pm\sqrt{\frac{b^2-4ac}{4a^2}} \\x&=&-\frac{b}{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}} \\&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{eqnarray}$$