\[ax^2+bx+c = 0 \]\[\\ \\ \]\[a(x^2+\frac{b}{a}x+\frac{c}{a}) = 0 \]\[\\ \\ \]\[x^2+\frac{b}{a}x = -\frac{c}{a} \]\[\\ \\ \]\[x^2+\frac{b}{a}x+(\frac{b}{2a})^2 = -\frac{c}{a}+(\frac{b}{2a})^2 \]\[\\ \\ \]\[(x+\frac{b}{2a})^2 = \frac{b^2-4ac}{4a^2} \]\[\\ \\ \]\[x_{1} = \frac{\sqrt[]{b^2-4ac}-b}{2a} \]\[\\ \\ \]\[x_{2} = \frac{-\sqrt[]{b^2-4ac}-b}{2a} \]\[\\ \\ \]\[设b^2-4ac=\Delta \]\[\\ \\ \]\[\because x_{1}=\frac{ \sqrt[]{\Delta}-b}{2a} \]\[\\ \\ \]\[\because x_{2}=\frac{ -\sqrt[]{\Delta}-b}{2a} \]\[\\ \\ \]\[\therefore x_{1}+x_{2}=-\frac{b}{a} \]\[\\ \\ \]\[\therefore x_{1} \cdot x_{2}=\frac{c}{a} \] 标签：frac,therefore,推导,定理,Delta,sqrt,韦达,4ac,2a From： https://www.cnblogs.com/Preparing/p/16755174.html