Solve the equation $\sqrt{2x+1} - \sqrt{x} = 1$. Express $\frac{\left(2\sqrt{3}-5\right)^2}{\sqrt{3}-2}$ in the form $p\sqrt{3} + q$ where $p$ and $q$ are integers.

1. Rearrange the equation to isolate one of the surds and square both sides.

$\sqrt{2x+1} = \sqrt{x} + 1$
$(\sqrt{2x+1})^2 = (\sqrt{x} + 1)^2$
$2x + 1 = x + 2\sqrt{x} + 1$

2. Simplify the equation to isolate the remaining surd.

$2x - x + 1 - 1 = 2\sqrt{x}$
$x = 2\sqrt{x}$

3. Square both sides again to remove the remaining surd and solve for $x$.

$x^2 = (2\sqrt{x})^2$
$x^2 = 4x$
$x^2 - 4x = 0$
$x(x - 4) = 0$
$x = 0 \text{ or } x = 4$

4. Check the solutions in the original equation to reject any extraneous roots.

Substitute $x = 0$:
LHS $= \sqrt{2(0)+1} - \sqrt{0} = 1 - 0 = 1 = \text{RHS}$
Substitute $x = 4$:
LHS $= \sqrt{2(4)+1} - \sqrt{4} = \sqrt{9} - 2 = 3 - 2 = 1 = \text{RHS}$
Both solutions are valid.

1. Expand the numerator

$\left(2\sqrt{3}-5\right)^2 = (2\sqrt{3})^2 - 2(2\sqrt{3})(5) + 5^2$
$= 12 - 20\sqrt{3} + 25$
$= 37 - 20\sqrt{3}$

2. Multiply numerator and denominator by the conjugate of the denominator

$\frac{37 - 20\sqrt{3}}{\sqrt{3}-2} = \frac{(37 - 20\sqrt{3})(\sqrt{3}+2)}{(\sqrt{3}-2)(\sqrt{3}+2)}$

3. Expand the new numerator and simplify the denominator

Denominator: $(\sqrt{3})^2 - 2^2 = 3 - 4 = -1$
Numerator: $37\sqrt{3} + 74 - 20(3) - 40\sqrt{3}$
$= 37\sqrt{3} + 74 - 60 - 40\sqrt{3}$
$= 14 - 3\sqrt{3}$

4. Simplify to the final form

$\frac{14 - 3\sqrt{3}}{-1} = -14 + 3\sqrt{3}$
$= 3\sqrt{3} - 14$