Answer
The solution set is $\left\{\right\}$ or $\emptyset$.
Work Step by Step
Square both sides to obtain:
$3-x+x^2=(x-2)^2$
Use the rule $(a-b)^2 = a^2-2ab+b^2$ to obtain:
$3-x+x^2=x^2-2(x)(2)+2^2
\\3-x+x^2=x^2-4x+4$
Put all terms on the left side of the equation, and then combine like terms.
Note that when a term is transferred to the other side of an equation, its sign/operation changes to its opposite.
$3-x+x^2-x^2+4x-4=0
\\(x^2-x^2)+(-x+4x) + (3-4) = 0
\\3x-1=0$
Add $1$ on both sides of the equation to obtain:
$3x=1$
Divide 3 on both sides of the equation to obtain:
$x=\dfrac{1}{3}$
However, note that when $\frac{1}{3}$ is substituted to $x$ in the equation, the right side becomes $\frac{1}{3} - 2 = -\frac{5}{3}$
A principal square root is never negative.
This means that $\frac{1}{3}$ is an extraneous solution.
Thus, the solution set is $\left\{\right\}$ or $\emptyset$.
