(2x^2 -3x -4)^2 - (x^2 - x)^2 = 0

$(2x^2-3x-4)^2-(x^2-x)^2=0$

$⇔(2x^2-3x-4+x^2-x)(2x^2-3x-4-x^2+x)=0$

$⇔(3x^2-4x-4)(x^2-2x-4)=0$

$⇔(3x^2+2x-6x-4)[(x-1)^2-5]=0$

$⇔(x-2)(3x+2)[(x-1)^2-5]=0$

$⇔$ \(\left[ \begin{array}{l}x=2\\x=-\dfrac{2}{3} \\ x=1-\sqrt{5} \\ x=\sqrt{5}+1 \end{array} \right.\) 

Vậy `S={2;-2/3;1-\sqrt{5};\sqrt{5}+1}`