Answer
The solution is $12$.
Work Step by Step
Subtract $6$ and add $\sqrt{3x}$ to both sides of the equation to obtain:
$$x-\sqrt{3x}-6+\sqrt{3x}=6-6+\sqrt{3x}
\\x-6=\sqrt{3x}$$
Square both sides. Note that $(a-b)^2=a^2-2ab+b^2$.
$$(x-6)^2=(\sqrt{3x})^2
\\x^2-2(x)(6)+6^2=3x
\\x^2-12x+36=3x
\\x^2-12x+36-3x=0
\\x^2-15x+36=0$$
Factor the trinomial:
$$(x-12)(x-3)=0$$
Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
\\&x-12=0 &\text{or} &x-3=0
\\&x=12 &\text{or} &x=3
\end{array}
Note that $3$ is does not satisfy the original equation so it is not a solution.
Therefore, the solution is $12$.
