Factor $12=2^{2}\times 3$. Rewrite the square root of the product $\sqrt{2^{2}\times 3}$ as the product of square roots $\sqrt{2^{2}}\sqrt{3}$. Take the square root of $2^{2}$.
$$\frac{6-\sqrt{3}}{9-2\sqrt{3}}$$
Rationalize the denominator of $\frac{6-\sqrt{3}}{9-2\sqrt{3}}$ by multiplying numerator and denominator by $9+2\sqrt{3}$.
Consider $\left(9-2\sqrt{3}\right)\left(9+2\sqrt{3}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.
