Answer
The solution set is $\left\{4\right\}$.
Work Step by Step
Subtract 2 on both sides of the equation to obtain:
$\sqrt{12-2x}=x-2$
Square both sides to obtain:
$12-2x=(x-2)^2$
Use the rule $(a-b)^2 = a^2-2ab+b^2$ to obtain:
$12-2x=x^2-2(x)(2) + 2^2
\\12-2x=x^2-4x+4$
Subtract $12$ and add $2x$ on both sides of the equation, and then combine like terms to obtain:
$\begin{array}{ccc}
&12-2x-12+2x &= &x^2-4x+4-12+2x
\\&0 &= &x^2-2x-8
\end{array}$
Factor the trinomial to obtain:
$0=(x-4)(x+2)$
Equate each factor to zero, and then solve each equation to obtain:
$\begin{array}{ccc}
&x-4=0 &\text{ or } &x+2=0
\\&x=4 &\text{ or } &x=-2
\end{array}$
$-2$ is an extraneous solution since it does not satisfy the original equation.
Thus, the solution set is $\left\{4\right\}$.
