How to find the $x^2+y^2=?$ Suppose that $ x$ and $ y$ are complex numbers, such $x+y=1$ and $x^{20}+y^{20}=20$, Find the sum of all possible values $x^2+y^2$ 
I find this problem answer is $-90$, But my methods is very very ugly, and I think this Problem have some nice methods.Thank you everyone.

Let $xy=c$
So, $x^2+y^2=(x+y)^2-2xy=1-2c$
$x^3+y^3=(x+y)^3-3xy(x+y)=1-3c$
On multiplication, $(x^2+y^2)(x^3+y^3)=(1-2c)(1-3c)$
$\implies x^5+y^5+x^2y^2(x+y)=1-5c+6c^2\implies x^5+y^5=1-5c+5c^2$
On squaring, $ x^{10}+y^{10}+2(xy)^5=(1-5c+5c^2)^2$
$\implies  x^{10}+y^{10}=1-10c+35c^2-50c^3+25c^4-2c^5$
On squaring, $ x^{20}+y^{20}+2(xy)^{10}=(1-10c+35c^2-50c^3+25c^4-2c^5)^2$
$\implies  x^{20}+y^{20}=4c^{10}-100c^9+\cdots+1-2c^{10}$
$\implies  2c^{10}-100c^9+\cdots+1=20\implies 2c^{10}-100c^9+\cdots-19=0$
This is a $10$ degree equation in $c=xy$
Using Vieta's formula, $\sum xy=\frac{100}2=50$
$\implies \sum x^2+y^2=10-2\sum xy=10-2\cdot 50=-90$
