Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I think that in the case $A=0$ one of the results of the two formulas should match $-\frac{C}{B}$. Yet, if $A=0$, $\frac{-B + \sqrt{B^2 - 4AC}}{2\times0}$, we will get $\frac{0}{0}$, and with $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$ we will get $-\frac{B}{0}$, so none of the two possible solutions will match $-\frac{C}{B}$. Shouldn't there be at least one solution that has the same value as $-\frac{C}{B}$?

You can use the conjugate quantity to get an alternative expression of the roots:
Suppose that $B > 0$. Then:
$\frac{-B + \sqrt{B^2-4AC}}{2A} = \frac{(-B + \sqrt{B^2-4AC})(-B - \sqrt{B^2-4AC})}{2A(-B - \sqrt{B^2-4AC})} = \frac{(-B)^2 - (\sqrt{B^2-4AC})^2}{2A(-B - \sqrt{B^2-4AC})} = \frac{2C}{-B - \sqrt{B^2-4AC}}$
If $A = 0$, the denominator is $-B - \sqrt{B^2} = -B - |B| = -2B$.
Therefore the root is $-\frac{C}{B}$.
