What will happen to the roots of $ax^2 + bx + c = 0$ if the $a \to 0$? Exercise:

What will happen to the roots of the quadratic equation
  $$ax^2 + bx + c = 0$$
  if the coefficient $a$ approaches zero while the coefficients $b$ and $c$ are constant, and $b \neq 0$?


Attempt:
$\lim\limits_{a \to 0}{(ax^2 + bx + c)} = bx + c = 0 \longrightarrow x = -\frac{c}{b}$
However, I don't think my solution is complete; shouldn't I end up with $2$ roots? (I've only found $1$.)

Request:
Is there indeed another root to find? If so, how to I find it?

If $b=0$, then the roots are (if real), $\pm\sqrt{-c/a}$. If $c\ne0$, both roots tend to infinity (positive and negative).
If $b\ne0$, it's not restrictive to assume $b>0$ (otherwise multiply by $-1$). 
If $c=0$, the roots are $0$ and $-b/a$, the latter tending to infinity ($\infty$ if $a$ approaches $0$ from the negative side, $-\infty$ if $a$ approaches $0$ from the positive side).
Assume $c\ne0$. The discriminant will be positive when $a$ belongs to a suitable (punctured) neighborhood of $0$, namely for $0<|a|<b^2/|4c|$, so we have two roots.
Now we can rationalize the expression for the roots.
First root:
$$
\frac{-b+\sqrt{b^2-4ac}}{2a}=
\frac{b^2-b^2+4ac}{2a(-b-\sqrt{b^2-4ac})}=
-\frac{2c}{\sqrt{b^2-4ac}+b}
$$
For $a\to0$ this has limit $-c/b$. This is to be expected, because the polynomial will become $bx+c$ for $a=0$.
Second root:
$$
\frac{-b-\sqrt{b^2-4ac}}{2a}=
\frac{b^2-b^2+4ac}{2a(-b+\sqrt{b^2-4ac})}=
\frac{2c}{\sqrt{b^2-4ac}-b}
$$
If $a\to0$, then this has limit $\pm\infty$ (according to the sign of $c$).

