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?