#### Answer

$x=2$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
\sqrt{x^2-3x+3}=x-1
,$ square both sides of the equation. Use special products and concepts of solving quadratic equations. Finally, do checking of the solution with the original equation.
$\bf{\text{Solution Details:}}$
Squaring both sides of the equation results to
\begin{array}{l}\require{cancel}
\left( \sqrt{x^2-3x+3} \right)^2=(x-1)^2
\\\\
x^2-3x+3=(x-1)^2
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x^2-3x+3=(x)^2-2(x)(1)+(1)^2
\\\\
x^2-3x+3=x^2-2x+1
.\end{array}
Using the properties of equality to isolate the variable results to
\begin{array}{l}\require{cancel}
(x^2-x^2)+(-3x+2x)=1-3
\\\\
-x=-2
\\\\
x=2
.\end{array}
Upon checking, $
x=2
$ satisfies the original equation.