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?

I think that my last question (in the intention of a 'slightly' different issue) accidentally can be helpful. It contains alternative expressions for the roots of the quadratic equation and their behavior for different ($\mathbb{R}$) coefficients.
Really universal quartic formula (Do not be afraid of the 'quartic' - there is more meaningful content about quadratic.)
I will present the reasoning for $\mathbb{R}$ coefficients, postponing their checking and possible generalization for the $\mathbb{C}$ ones.
We know that in the case of a quadratic equation $ax^2+bx+c=0$, there are two alternative formulas for (both) roots:
$$x_{ \begin{array}{c}
1 \\ 
2 \end{array}
}=\frac{-b\mp \sqrt{\mathrm{\Delta }}}{2a}$$
and
$$x_{ \begin{array}{c}
1 \\ 
2 \end{array}
}=\frac{2c}{-b\mp \sqrt{\mathrm{\Delta }}}$$ 
The second one can be obtained from the first by expanding the fraction with an complementary expression to the difference of squares.
$$x_{ \begin{array}{c}
1 \\ 
2 \end{array}
}=\frac{-b\mp \sqrt{\mathrm{\Delta }}}{2a}\cdot \frac{-b\pm \sqrt{\mathrm{\Delta }}}{-b\pm \sqrt{\mathrm{\Delta }}}=\frac{2c}{-b\pm \sqrt{\mathrm{\Delta }}}$$ 
or also by using of the product Vieta formula.
$$x_{ \begin{array}{c}
1 \\ 
2 \end{array}
}\cdot x'_{ \begin{array}{c}
2 \\ 
1 \end{array}
}=\frac{c}{a}$$ 
$$x'_{ \begin{array}{c}
2 \\ 
1 \end{array}
}=\frac{c}{a}\cdot \frac{2a}{-b\mp \sqrt{\mathrm{\Delta }}}=\frac{2c}{-b\mp \sqrt{\mathrm{\Delta }}}$$ 
 (In the second method, the formulas for both roots were found in the reverse order, but it does not matter.)
The first one works well $\forall b,c$ and $\forall a\neq 0$, the second one $\forall a,b$ but $\forall c\neq 0$.
In the first case, you can also proceed by calculating the appropriate limits. Then one of the roots will go to infinity, and the other will go smoothly into the only element of the obtained linear equation.
$$x_1=\frac{-b-\sqrt{\mathrm{\Delta }}}{2a}{{\mathop{\longrightarrow}\limits_{a\to 0}}}\left\{ \begin{array}{cc}
\mathrm{for}\ b<0: & -\frac{c}{b} \\ 
\mathrm{for}\ b=0: & \infty \left[-\sqrt{-{\mathrm{sgn} c\ }\ }\right] \\ 
\mathrm{for}\ b>0: & -\infty  \end{array}
\right.$$ 
$$x_2=\frac{-b+\sqrt{\mathrm{\Delta }}}{2a}{{\mathop{\longrightarrow}\limits_{a\to 0}}}\left\{ \begin{array}{cc}
\mathrm{for}\ b<0: & \infty  \\ 
\mathrm{for}\ b=0: & \infty \left[\sqrt{-{\mathrm{sgn} c\ }\ }\right] \\ 
\mathrm{for}\ b>0: & -\frac{c}{b} \end{array}
\right.$$ 
Wherein, only finite results make sense as solutions to the linear equation!
Which root of the quadratic equation will go into the root of the linear equation depends on the sign of the linear coefficient $b$.
Formula 
$x_{ \begin{array}{c}
1 \\ 
2 \end{array}
}=\frac{2c}{-b\mp \sqrt{\mathrm{\Delta }}}$ 
reproduces these results without the use of limits, but it itself requires moving to the limit when $c\ =\ 0$.
$$x_1=\frac{2c}{-b+\sqrt{\mathrm{\Delta }}}{{\mathop{\longrightarrow}\limits_{c\to 0}}}\left\{ \begin{array}{cc}
\mathrm{for}\ b<0: & -\frac{b}{a} \\ 
\mathrm{for}\ b=0: & 0 \\ 
\mathrm{for}\ b>0: & 0 \end{array}
\right.$$ 
$$x_2=\frac{2c}{-b-\sqrt{\mathrm{\Delta }}}{{\mathop{\longrightarrow}\limits_{c\to 0}}}\left\{ \begin{array}{cc}
\mathrm{for}\ b<0: & 0 \\ 
\mathrm{for}\ b=0: & 0 \\ 
\mathrm{for}\ b>0: & -\frac{b}{a} \end{array}
\right.$$ 
Here all the results have meaning.
That which root of the quadratic equation will go to $0$ and which in a potentially non-zero root of the obtained equation depends on the sign of the linear coefficient $b$.
There is also the possibility of refining formulas so that the root that can be calculated directly (without the use of limits) is always transformed into the root of the obtained equation.
$$x_1=\frac{-b-{\mathrm{sgn} b\ }\sqrt{\mathrm{\Delta }}}{2a}={{\left|c=0\right|}}=-\frac{b}{a}$$ 
$$x_2=\frac{2c}{-b-{\mathrm{sgn} b\ }\sqrt{\mathrm{\Delta }}}={{\left|a=0\right|=}}\left\{ \begin{array}{cc}
\mathrm{for}\ b\neq 0: & -\frac{c}{b} \\ 
\mathrm{for}\ b=0: & ? \end{array}
\right.$$ 
Of course, only the case $b\ \neq \ 0$ have meaning.
The most universal (?) set of quadratic formulas is then:
$$x_1=\frac{-b-{\mathrm{sgn} b\ }\sqrt{\mathrm{\Delta }}}{2a}$$ 
$$x_2=\frac{2c}{-b-{\mathrm{sgn} b\ }\sqrt{\mathrm{\Delta }}}$$
and they work for all $\mathbb{R}$ values of coefficients. (I leave the generalization for $\mathbb{C}$ numbers for later.)
