$\begin{aligned}A \rightarrow B + C\end{aligned}$

The initial amount and the change:

$\begin{aligned}A,B,C &= a,b,c\\ &\rightarrow -y,+y,+y\end{aligned}$

Equilibrium

$\begin{aligned}k = \frac{(b + y)(c + y )}{a - y}\\0 = y^2 + (b + c + k) y + b c - k a\\y= \frac{- (b + c + k) + \sqrt{(b + c + k)^2 - 4 b c + 4 k a}}{2} \end{aligned}$

If negative square root is accepted, by $c+y>0$

$\begin{aligned}0 < c + \frac{- (b + c + k) - \sqrt{(b + c + k)^2 - 4 b c + 4 k a}}{2}\\\sqrt{(b+c+k)^2-4b c+4k a}<c-b-k\\(b+c+k)^2-4b c+4k a<(c-b-k)^2\\(b+c+k)^2-(c-b-k)^2<4b c-4k a\\2c(2b+2k)<4b c-4k a\\c<-a\end{aligned}$

Similarly, $b < -a$

So, $\max(b, c)< -a$

If a contradiction occurs, then the one with negative squared root shall be dropped. A contradiction if one of $b$ and $c$ are non-negative and $a$ is positive.

Another explanation as below.

With the equation $0 = A x^2 + B x + C, A>0$ and the requirement that the solution shall accommodate the case when $A$ is tiny around zero, we have

$x = \frac{-B -\sqrt{B^2 - 4 A C}}{2 A}$$x =\frac{-B + \sqrt{B^2 - 4 A C}}{2 A}$

The solution must take limit to be the right answer $x = \frac{-C}{B}$ when $A$ takes limit to zero

Therefore it must be:

- When $B$ is positive, the right answer is positive square root - When $B$ is negative, the right answer is negative square root

And the original equation is in fact

$\begin{aligned}0 = \frac{1}{b + c + k}y^2 + y + \frac{b c - k a}{b + c + k}\end{aligned}$

whose $B$ is positive