\frac{2}{3n}-\frac{2}{3}=\frac{n}{6}+\frac{4}{3} \ |* 6n , 3n\ne 0\Rightarrow n\ne 0
\frac{12n}{3n}-\frac{12n}{3}=\frac{6n^2}{6}+\frac{24n}{3}
4-4n=n^2+8n
4-4n-n^2-8n=0
-n^2-12n+4=0
a=-1, b=-12, c=4
\Delta=b^2-4ac=(-12)^2-4\cdot (-1)\cdot 4=144+16=160
\sqrt\Delta=\sqrt{160}=\sqrt{16\cdot 10}=4\sqrt{10}
n_1=\frac{-b-\sqrt\Delta}{2a}=\frac{12-4\sqrt{10}}{2\cdot (-1)}=\frac{12-4\sqrt{10}}{-2}=-6+2\sqrt{10}=2(\sqrt{10}-3)
n_2=\frac{-b+\sqrt\Delta}{2a}=\frac{12+4\sqrt{10}}{2\cdot (-1)}=\frac{12+4\sqrt{10}}{-2}=-6-2\sqrt{10}=-2(3+\sqrt{10})