ax2+bx+c=0 (a0)x2+bax+ca=0x2+2b2ax=cax2+2b2ax+b24a2=b24a2cax2+2b2ax+b24a2=b24ac4a2(x+b2a)2=b24ac4a2x+b2a=±b24ac4a2x=±b24ac4a2b2ax=±b24ac2ab2ax=±b24acb2a\begin{aligned} ax^2 + bx + c &= 0 \ (a \ne 0 )\\ x^2 + \frac bax + \frac ca &= 0\\ x^2 + 2\frac b{2a}x &= -\frac ca\\ x^2 + 2\frac b{2a}x + \frac{b^2}{4a^2} &= \frac{b^2}{4a^2} - \frac ca\\ x^2 + 2\frac b{2a}x + \frac{b^2}{4a^2} &= \frac{b^2 - 4ac}{4a^2}\\ \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 &= \pm\sqrt\frac{b^2 - 4ac}{4a^2} - \frac b{2a}\\ x &= \pm\frac{\sqrt{b^2 - 4ac}}{2a} - \frac b{2a}\\ x &= \frac{\pm\sqrt{b^2 - 4ac} - b}{2a}\\ \end{aligned}

如果 b2<4acb^2 < 4ac,则方程没有实数解。如果 b2=4acb^2 = 4ac,则方程有两个相同的实数解。如果 b2>4acb^2 > 4ac,则方程有两个不同的实数解。