一元二次方程式求解

$(ax^{2}+bx+c=0)$

二邊除a : $(x^{2}+\frac{bx}{a}+\frac{c}{a}=0)$

$(\therefore x^{2}+\frac{bx}{a}=-\frac{c}{a})$

二邊加上$((\frac{b}{2a})^{2})$  => $(x^{2}+\frac{bx}{a}+(\frac{b}{2a})^{2}=-\frac{c}{a}+(\frac{b}{2a})^{2})$

$(\therefore (x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}})$

$(x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a})$