Infinite Equation

Nevermind found i!. It's $\sqrt{2}$

Find:

$x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.$

$\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}} \\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\ x-1 &=& \cfrac{1}{2 + x-1} \\ (x-1)(2 + x-1) &=& 1 \\ x+x^2-x-2-x+1 &=& 1 \\ x^2-2 &=& 0 \\ x^2 &=& 2 \\ \mathbf{ x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array}$