Question: Solve: $\dfrac{3x-4}{2+ \sqrt{3x}} - \dfrac{\sqrt{3x} -2 }{2} = 2$

Solution:
Given,

$\dfrac{3x-4}{2+ \sqrt{3x}} - \dfrac{\sqrt{3x} -2 }{2} = 2$

$or, \dfrac{(\sqrt{3x})^2 - 2^2}{\sqrt{3x} + 2}  - \dfrac{\sqrt{3x} -2 }{2} = 2$

$or, \dfrac{(\sqrt{3x} +2)(\sqrt{3x} - 2)}{\sqrt{3x} + 2}  - \dfrac{\sqrt{3x} -2 }{2} = 2$

$or, (\sqrt{3x} -2 )  - \dfrac{\sqrt{3x} -2 }{2} = 2$

$or, \dfrac{2(\sqrt{3x} - 2) - (\sqrt{3x} -2 ) }{2} = 2$

$or, 2\sqrt{3x} - 4 - \sqrt{3x} + 2 = 2*2$

$or, \sqrt{3x} -2 = 4$

$or, \sqrt{3x} = 4+2$

$or, \sqrt{3x} = 6$

squaring both sides

$or, (\sqrt{3x})^2 = 6^2$

$or, 3x = 36$

$or, x = \dfrac{36}{3}$

$\therefore x= 12$