Find $x,y \in \mathbb C $ such that $x^5+y^5=275, x+y=5$. My attempt:
Let $x, y$ be roots of
$$t^2-5t+p=0$$
I got
$$p=\frac{5 \pm \sqrt{-431}}{2}$$
using Vieta's relations.
Now I just calculated x and y using the quadratic formula in $t^2-5t+p=0$.
But I got two very hideous complex values of x and y. So I think my answer may be wrong.
The values I got are
$$x= \frac{5+\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}$$ or
$$x= \frac{5+\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}$$
Please post your own solutions too if you have any different solutions.

Hint: Substituting $y=5-x$ the equation $x^5+y^5=275$ is equivalent to
$$
(x^2 - 5x + 19)(x - 2)(x - 3)=0
$$
