Basic calculus is the easy way: the second derivative is negative.
If you want to use the formal definition, there is a bit less hassle if you note that $-x^2+2x=-(x-1)^2+1$. Now make the change of variable $w=x-1$, just a shift, does not affect the geometry. So we are looking at the function $-w^2+1$. Pull it down by $1$, does not affect the geometry. We are trying to prove that $-w^2$ is concave. I don't like negative numbers much, flip sign. So we want to prove that $w^2$ is convex.
Let $0 \le t\le 1$. We want to show that $(tx+(1-t)y)^2 \le tx^2+(1-t)y^2$. Expand the left-hand side. We want to show that
$tx^2+(1-t)y^2-t^2x^2-2t(1-t)xy-(1-t)^2y^2 \ge 0$.
A bit of algebra reduces the above expression to
It is not difficult to show this is non-negative!