That is much better. And no the answer is not 6.
\(x(3x-4) \le \frac{(6x^2-3x+5)}{10}\\ 10x(3x-4) \le 6x^2-3x+5\\ \)
Now you have to expand the left-hand side
then
take all the terms to the left side so that it is <=0
then
Factorise the left side.
Now if you let the left side equal y
then y must be less than or equal to 0
The left side, when you put it equal to y, will actually be a concave up parabola.
Find the roots of the parabola.
Any integers between the roots will be solutions to this problem. :)
Give it a go 