Answer
See derivation below.
The minus sign yields a negative value for $e^{y}$, so it is ommited from being a part of the solution.
Work Step by Step
$\sinh^{-1}x=y\quad$ means that $x=\sinh y$
By definition of sinh:
$ x=\displaystyle \frac{e^{y}-e^{-y}}{2}\quad $... we solve this for y
$2x=e^{y}-\displaystyle \frac{1}{e^{y}}\qquad $... substitute $t=e^{y} $ (we see that $t\gt 0)$
$ 0=t-\displaystyle \frac{1}{t}-2x\qquad$ ... multiply with t
$t^{2}-(2x)t-1=0\qquad $... apply the quadratic formula, $a=1, b=-2x, c=-1$
$t=\displaystyle \frac{2x\pm\sqrt{4x^{2}+4}}{2}=\frac{2x\pm 2\sqrt{x^{2}+1}}{2}=x\pm\sqrt{x^{2}+1}$
$\sqrt{x^{2}+1}\gt\sqrt{x^{2}}\quad$
so we will discard the minus, because it yields a negative $t$ ($t$ needs to be positive).
$t=x+\sqrt{x^{2}+1}$
$e^{y}=x+\sqrt{x^{2}+1}\qquad $... apply ln(..) to both sides
$y=\ln(x+\sqrt{x^{2}+1})$
