The problem is to simplify the expression $\frac{4}{2\sqrt{3}-3}$. This involves rationalizing the denominator.

1. Problem Description

The problem is to simplify the expression $\frac{4}{2\sqrt{3}-3}$. This involves rationalizing the denominator.

2. Solution Steps

To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of $2\sqrt{3}-3$ is $2\sqrt{3}+3$.

$\frac{4}{2\sqrt{3}-3} = \frac{4}{2\sqrt{3}-3} \cdot \frac{2\sqrt{3}+3}{2\sqrt{3}+3}$

$= \frac{4(2\sqrt{3}+3)}{(2\sqrt{3}-3)(2\sqrt{3}+3)}$

We need to multiply out the denominator:

$(2\sqrt{3}-3)(2\sqrt{3}+3) = (2\sqrt{3})^2 - (3)^2 = 4(3) - 9 = 12-9=3$

So, the expression becomes:

$\frac{4(2\sqrt{3}+3)}{3} = \frac{8\sqrt{3}+12}{3}$

3. Final Answer

$\frac{8\sqrt{3}+12}{3}$