Add up the first and the last to get $\textstyle \sum_i x_i = \textstyle \sum_i x_i^2 + 6x_1x_2+6x_3x_4$. So, for any permutation $\sigma:\{1,\dots,4\}\to\{1,\dots,4\}$, $x_{\sigma(1)}x_{\sigma(2)} = x_{\sigma(3)}x_{\sigma(4)}$. Now using $x_1x_2 + x_3x_4 = x_1x_4 + x_2x_3$, we get $(x_1-x_4)(x_2-x_3)=0$. So either $x_1=x_4$ or $x_2=x_3$. Assume the former. Now inspecting $x_1x_2 +x_3x_4 = x_1x_4+x_2x_3$ we get $(x_1-x_3)(x_2-x_4)=0$ so either $x_1=x_3=x_4$ or $x_2=x_3=x_4$. Namely, at least three of the variables are equal. Moreover, adding all up we get $3(x_1+x_2+x_3+x_4) = 3(x_1+x_2+x_3+x_4)^2$, so $x_1+x_2+x_3+x_4\in\{0,1\}$. From here, through symmetry, one has to investigate two cases: $(x_1,x_2,x_3,x_4)=(c,c,c,-3c)$ and $(x_1,x_2,x_3,x_4)=(c,c,c,1-3c)$. The rest is routine and omitted.