These roots are $a-3b,a-b,a+b,a+3b$ for some real $a,b$ We have $x_1+x_2+x_3+x_4=0 = 4a \to a=0$ $-2(3n+2)=(x_1+x_2+x_3+x_4)^2-(x_1^2+x_2^2+x_3^2+x_4^2)=-20b^2 \to 10b^2=3n+2 \to n=\frac{10b^2-2}{3}$ $n^2=x_1x_2x_3x_4=9b^4 \to n= \pm 3b^2$ So $ \frac{10b^2-2}{3}=3b^2 \to b^2=2,n=6$ or $\frac{10b^2-2}{3}=-3b^2,b^2=\frac{2}{19},n=-\frac{6}{19}$

Since the sum of the roots is $0$, they must be the progression $-3r,-r,r,3r$ for some $r\in\mathbb{R}^+$. $(x-3r)(x-r)(x+r)(x+3r)=(x^2-9r^2)(x^2-r^2)=x^4-10rx^2+9r^4$ So we are solving the system: $3n+2=10r^2$ $n^2=9r^4$ The second equation gives us $3r^2=\pm n$ so putting it in the first equation gives $9n+6=\pm 10n$ $9n+6=\pm 10n\implies(9\pm 10)n=-6\implies n\in\left\{6,-\frac{6}{19}\right\}$. We must check that the roots are real because you may get a complex arithmetic progression $-3bi,-bi,bi,3bi$ with common difference $2bi$. $n=6\implies r^2=\frac{3n+2}{10}=2>0$. Good. $n=-\frac{6}{19}\implies r^2=\frac{3n+2}{10}=\frac{2}{19}>0$. Good.