#### Answer

$-i\pm i\sqrt{1-i}$

#### Work Step by Step

$i{{x}^{2}}-2x+1=0$
Now compare the equation $i{{x}^{2}}-2x+1=0$ with the standard form of the quadratic equation $a{{x}^{2}}+bx+c=0$.
Here,
$a=i,b=-2$and $c=1$
Substitute $a=i,b=-2$ and $c=1$ into the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
$\begin{align}
& x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( i \right)\left( 1 \right)}}{2\left( i \right)} \\
& =\frac{2\pm \sqrt{4-4i}}{2i} \\
& =\frac{2\pm 2\sqrt{1-i}}{2i} \\
& =\frac{1\pm \sqrt{1-i}}{i}
\end{align}$
Rationalize the expression by multiplying by $\frac{i}{i}$,
$\begin{align}
& x=\frac{1\pm \sqrt{1-i}}{i}\cdot \frac{i}{i} \\
& =\frac{i\pm i\sqrt{1-i}}{{{i}^{2}}} \\
& =\frac{i\pm i\sqrt{1-i}}{-1} \\
& =-i\pm i\sqrt{1-i}
\end{align}$