Why do we get two solutions when inverting $y = \sinh x$? Using the definition $\sinh x = \dfrac{e^x-e^{-x}}{2},\;$ let's say we want to solve $\;y = \sinh x \;$ for $x$. 
It's not hard to show that $\;\sinh x \;$ is bijective, so this should have exactly one solution:
$$\begin{align}
y &= \frac{e^x-e^{-x}}{2} \\ \\
y &= e^x - e^{-x} \\ \\
2ye^x &= (e^x)^2 - 1 \\
\end{align}$$
$$(e^x)^2 - 2y e^x -1 = 0$$
$$ e^x = y \pm \sqrt{1+y^2} $$
Now, the solution with "$-$" cannot be, since it's always negative and we can't take the log. Therefore, we get that $y = \sinh x \iff x = \ln(y+\sqrt{1+y^2})$.
What I don't understand is, why are there two solutions to the quadratic? You would think that since we start with the equation for a bijective function and apply reversible operations to it (multiplication by $2$, multiplication by $e^x$, subtracting $2ye^x$) we would get something that has only one solution, but somehow we get two and one has to be discarded. Why does that happen?

Here is a diametrically opposite interpretation from that of the first (and until now the only) answer.
The method used is correct; solving the equation one shows that $e^x$ has one of two possible real values.  If a solution exists, then it is in such-and-such subset of $\mathbb{R}$.  That is a correct deduction.
This correct application of the quadratic equation is insufficient to isolate a unique $x$.  This may be surprising because $\sinh$ is injective on the reals.  But $\sinh$ is not uniquely invertible on the complex numbers, or matrices, other domains extending the reals but satisfying the same algebraic laws used in solving the quadratic equation.   Thus, the assumption that $x$ is real has to also be used as a separate piece of argument, and that means $e^x$ is positive.  This pins down a single value of $x$.
The algebra does not "know" that $x$ was restricted to $\mathbb{R}$, but that is not a problem with the algebra.
