Problem 530

Answer: $\{0.5 ;-\sqrt{505}\}$.

## Solution:

1) Let $x \geq 0$, then $(x-x)^{2}+2020(x+x)=2020$, we get $x=0.5$.
2) Let $x<0, \quad$ then $(x+x)^{2}+2020(x-x)=2020$, we get $4 x^{2}=2020, \quad x=$ $\pm \sqrt{505}$, but since $x<0$, then $x=-\sqrt{505}$.