Find all integers $x$ such that $\frac{2x^2-3}{3x-1}$ is an integer. Find all integers $x$ such that $\frac{2x^2-3}{3x-1}$ is an integer. 
Well if this is an integer then $3x-1 \mid 2x^2-3$ so $2x^2-3=3x-1(k)$ such that $k\in \mathbb{Z}$ from here not sure where to go I know that it has no solutions I just can't see the contradiction yet. 

Let $k=3x-1$ then $$3x\equiv_k 1$$ and $$2x^2-3 \equiv _k0$$ Multiplying last equation with 9 we get $$0\equiv _k2(3x)^2-27\equiv _k -25$$ So $$3x-1\mid 25 \implies 3x-1\in\{\pm 1,\pm5,\pm 25\}$$ and thus $$x\in\{0,2,-8\}$$
