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