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 $-\frac{C}{B}$?

There is an assumption in the derivation of the quadratic formula that $A \ne 0$. You can see it does not directly apply when this assumption is violated due to the $A$ in the denominator.
However, the solution of the linear case can nevertheless be obtained from the quadratic formula on the understanding one of the two solutions it yields is not valid.
First consider $A$ is very small, apply the binomial theorem to the square root
$$
\frac{-B\pm\sqrt{B^2-4AC}}{2A} \\
=\frac{-B\pm B\sqrt{1-4AC/B^2}}{2A} \\
\approx \frac{-B\pm B(1-2AC/B^2)}{2A} \\
= -B/2A\pm B/2A \mp C/B
$$
So far we still have 2 solutions. When $A=0$ we have a linear equation which has only has one solution, and one of the solutions from the quadratic formula is not valid.
Retaining the $+,-$ combination as the one valid solution,
$$
= -B/2A + B/2A - C/B \\
= -C/B.
$$
Note, there's also an assumption $B \ne 0$.
