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Let $a$ and $b$ be the roots of this equation. Then we want $$|a-b|=a^2+b^2.$$Squaring both sides, we get $$(a-b{)}^2=(a^2+b^2{)}^2.$$By Vieta's formulas, $a+b=-\frac{4}{5}$ and $ab=\frac{k}{5}.$ Squaring the equation $a+b=-\frac{4}{5},$ we get $$a^2+2ab+b^2=\frac{16}{25}.$$Then $$(a-b{)}^2=a^2-2ab+b^2=(a+b{)}^2-4ab=\frac{16}{25}-\frac{4k}{5}=\frac{16-20k}{25}.$$Also, $$a^2+b^2=\frac{16}{25}-2ab=\frac{16}{25}-\frac{2k}{5}=\frac{16-10k}{25}.$$Hence, $$\frac{16-20k}{25}={\left(\frac{16-10k}{25}\right)}^2.$$This simplifies to $25k^2+45k-36=0,$ which factors as $(5k-3)(5k+12)=0.$ Thus, the possible values of $k$ are $\boxed{\frac{3}{5},-\frac{12}{5}}.$