Answer
$(1,1);(-2,-2)$
Work Step by Step
$$\begin{cases} x^2+y=0\\x-y=2 \end{cases}$$
Add both equations:
$$x^2+y+x-y=0+2$$
$$x^2+x-2=0$$
$$(x+2)(x-1)=0$$
$$x=-2~~~or~~~x=1$$
Find corresponding $y$ values for each $x$ values:
$$-2-y=0\to y=--2$$
$$1-y=0\to y=1$$
Solutions are:
$$(1,1);(-2,-2)$$
