Sum of floor equation

For $1\le j\le n$ , we have $(2n^2+8)x^2+(2\sqrt{2}j-1)x+6j^2-2j\ge 6j^2-2j-\frac{(2\sqrt{2}j-1)^2}{4(2n^2+8)}\ge 6j^2-2j-1\ge 3$ . Then $\displaystyle 19n-16=\sum_{j=1}^{n}\lfloor{ (2n^2+8)x^2+(2\sqrt{2}j-1)x+6j^2-2j} \rfloor \ge \sum_{j=1}^{n}(6j^2-2j-1)=6\cdot\frac{n(n+1)(2n+1)}{6}-n(n+1)-n=2n^2(n+1)-n$ ,

which implies $(n-1)(n-2)(n+4)\le 0$ .Therefore the only possible n's are 1 and 2.

For $n=1,x=-0.1$ is a solution. For $n=2,x=-0.1$ is a solution.

So the answer is $1+2=3$ .