ab^{2}+c\right) =4a^{2}b^{2}-4a^{2}b^{2}-4ac =-4ac. \end{eqnarray*}$$ Therefore $c=0$, implies $\Delta =0$. If $a=0$ see 1. -- $^{1}$The quadratic equation $$ Ax^{2}+Bx+C=0 $$ can be solved by the quadratic formula $$ x=\frac{-B\pm \sqrt{B^{2}-4AC}}{2A}=\frac{-B\pm \sqrt{\Delta }}{2A}. $$ If $\Delta =0$, the only root is $x_{1}=x_{2}=\frac{-B}{2A}$. In the present case $A=a,B=-2ab$ and $C=ab^{2}+c$. share | improve this