Solve the equation.
$$x-\sqrt{12-4x}=0$$

Step 1 ：First, we note that $x$ must be less than or equal to $3$, since $\sqrt{12-4x}$ is undefined if $x>3$.

Step 2 ：Then, we simplify the equation by moving $\sqrt{12-4x}$ to the right side of the equation, which gives $x=\sqrt{12-4x}$.

Step 3 ：Next, we square both sides of the equation to eliminate the square root, which gives $x^2=12-4x$.

Step 4 ：Rearranging the equation gives $x^2+4x-12=0$.

Step 5 ：Then, we factor the quadratic equation, which gives $(x+6)(x-2)=0$.

Step 6 ：Setting each factor equal to zero gives the solutions $x=-6$ and $x=2$.

Step 7 ：However, we note that $x=-6$ is not a valid solution, since it would result in a negative number under the square root in the original equation.

Step 8 ：Therefore, the only solution is $x=\boxed{{\displaystyle 2}}$.