1. The problem is to simplify the expression $$\frac{2}{\sqrt{3}-1}$$ so that there is no square root in the denominator. 2. To remove the square root from the denominator, we multiply both numerator and denominator by the conjugate of the denominator. The conjugate of $$\sqrt{3}-1$$ is $$\sqrt{3}+1$$. 3. Multiply numerator and denominator: $$\frac{2}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$$ 4. Simplify the denominator using the difference of squares formula: $$(\sqrt{3})^2 - (1)^2 = 3 - 1 = 2$$ 5. So the expression becomes: $$\frac{2(\sqrt{3}+1)}{2}$$ 6. Cancel the common factor 2 in numerator and denominator: $$\frac{\cancel{2}(\sqrt{3}+1)}{\cancel{2}} = \sqrt{3}+1$$ 7. Final answer: $$\sqrt{3}+1$$