how to find x in the question Sinh(x-3)=1 even if i expand it into e terms i cant seem to get it

How to find x in the question Sinh(x-3)=1 even if i expand it into e terms i cant seem to get it

I.

$$\small{\text{
$
\begin{array}{rcl}
\sinh{(x-3)} &=& 1 \quad | \quad \sinh^{-1} \\
x-3 &=& \sinh^{-1}(1) \\
x &=& 3 + \sinh^{-1}(1) \\
x &=& 3 + 0.88137358702\\
x &=& 3.88137358702\\
\end{array}
$
}}$$

II.

$$\boxed{
\small{\text{
$
\sinh(x)=\frac{1}{2}\cdot \left( e^x - e^{-x}\right)
\qquad \sinh(x-3)=\frac{1}{2}\cdot \left( e^{x-3} - e^{-(x-3)}\right)
$
}}}$$

$$\small{\text{$
\begin{array}{rcl}
\sinh(x-3) &=& 1\\
\frac{1}{2}\cdot \left( e^{x-3} - e^{-(x-3)}\right) &=& 1\\
e^{x-3} - e^{-(x-3)} &=& 2 \\
e^{x-3} - \frac{1}{e^{x-3}} &=& 2 \quad | \quad u = e^{x-3}\\
u-\frac{1}{u} &=& 2 \quad | \quad \cdot u\\
u^2-1 &=& 2\cdot u \\
u^2 -2\cdot u - 1 &=& 0 \\
u_{1,2} &=& \frac{ 2\pm \sqrt{4-4\cdot(-1) } }{2}\\
u_{1,2} &=& \frac{ 2\pm \sqrt{2\cdot4 } }{2}\\
u_{1,2} &=& \frac{ 2\pm 2\cdot\sqrt{ 2 } }{2}\\
u_{1,2} &=& 1\pm \sqrt{ 2 }\\
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcll}
u &=& e^{x-3} \quad & | \quad \ln\\
\ln(u) &=& (x-3) \cdot \ln(e) \quad & | \quad \ln(e) = 1 \\
\ln(u) &=& x-3 \\
\boxed{x = 3 + \ln(u)}
\end{array}
$
}}$$

$$\small{\text{$
\begin{array}{rcl|rcl}
u_1 &=& 1+\sqrt{2} \quad & \quad u_2 &=& 1-\sqrt{2}\\
x_1 &=& 3 + \ln(u_1) \quad & \quad x_2 &=&3 + \ln(u_2) \\
x_1 &=& 3 + \ln(1+\sqrt(2)) \quad & \quad x_2 &=&3 + \ln(\underbrace{1-\sqrt(2)}_{<0\text{ no solution!}}) \\
x &=& 3 + \ln(1+\sqrt(2)) \\
x &=& 3 + \ln( 2.41421356237 ) \\
x &=& 3 + 0.88137358702\\
x &=& 3.88137358702\\
\end{array}
$
}}$$