The problem asks us to find the value of $t$ such that $\frac{2+\sqrt{3}}{1-\sqrt{t}} = \frac{-4}{2-\sqrt{3}}$.

1. Problem Description

The problem asks us to find the value of $t$ such that $\frac{2+\sqrt{3}}{1-\sqrt{t}} = \frac{-4}{2-\sqrt{3}}$.

2. Solution Steps

We have the equation $\frac{2+\sqrt{3}}{1-\sqrt{t}} = \frac{-4}{2-\sqrt{3}}$. First, we can cross-multiply:

$(2+\sqrt{3})(2-\sqrt{3}) = -4(1-\sqrt{t})$

The left side simplifies to:

$(2+\sqrt{3})(2-\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$

Thus, we have:

$1 = -4(1-\sqrt{t})$

$1 = -4 + 4\sqrt{t}$

$5 = 4\sqrt{t}$

$\sqrt{t} = \frac{5}{4}$

To find $t$, we square both sides:

$t = (\frac{5}{4})^2 = \frac{25}{16}$

3. Final Answer

The value of $t$ is $\frac{25}{16}$.