Find $\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}, x \gt 4$ 
Find
  $$\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}},  x \gt 4$$


My attempt :-
$$3(x-1)+2\sqrt{2x^2 - 7x - 4}$$ 
$$\implies 3x - 3 + 2\sqrt{2x^2 - 7x - 4} + 2x^2 - 2x^2 -7x + 7x - 1 +1 $$
$$\implies 2x^2 -7x  - 4 + 2\sqrt{2x^2 - 7x - 4} + (- 2x^2  + 10x  + 1) $$
let $y =  \sqrt{2x^2 - 7x - 4}$
$$\therefore y^2 + 2y + (- 2x^2  + 10x  + 1)$$
solving for $y$ :-
$$y = {-2\pm \sqrt{4 - 4\times(- 2x^2  + 10x  + 1)} \over 2} $$
$$\implies y = {-1\pm \sqrt{2x^2  - 5x}} $$

$$\therefore y^2 + 2y + (- 2x^2  + 10x  + 1) = (y + 1 + \sqrt{2x^2  - 5x})(y + 1 - \sqrt{2x^2  - 5x})$$
$$\therefore \sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}} = \sqrt{(y + 1 + \sqrt{2x^2  - 5x})(y + 1 - \sqrt{2x^2  - 5x})\over 2}$$
Sadly this does not remove the square root. 

So the question is how can i factorize $3(x-1)+2\sqrt{2x^2 - 7x - 4}$ to remove that square root ?
just some hints are fine with me, thanks. 

Hint: $2 x^2 - 7 x - 4 = (2x+1)(x-4)$, so you may guess that the desired result is of the form $a \sqrt{2x+1} + b \sqrt{x-4}$.  Now square that, and see what $a$ and $b$ would give you $\dfrac{3}{2}(x-1) + \sqrt{2x^2-7x-4}$.
