\begin{align*} &ax^2+bx+c=0 \ (a \neq 0) \\ &x^2+\frac{b}{a}x+\frac{c}{a}=0\\&(x+\frac{b}{2a})^2=-\frac{c}{a}+\frac{b^2}{4a^2}\\ &(x+\frac{b}{2a})^2=\frac{-4ac+b^2}{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{-b\pm\sqrt{b^2-4ac}}{2a} \end{align*}
Therefore,
If b^2-4ac>0, there are two real roots
If b^2-4ac=0, there is only one real root (or two repeated roots)
If b^2-4ac<0, there is no real root (or two complex conjugate root)