Answer
$(-3,-1)$ and $(1,3)$
Work Step by Step
$\left\{\begin{aligned}x^{2}+y^{2}&=10\\y&=x+2\end{aligned}\right.$
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Substitute y into the second equation
$\begin{aligned}x^{2}+(x+2)^{2}&=10\\x^{2}+x^{2}+4x+4&=10\\2x^{2}+4x-6&=0\\x^{2}+2x-3&=0\end{aligned}$
... two factors of $-3$ whose sum is $+2$ ... are $-1$ and $+3$.
$(x+3)(x-1)=0$
Back-substituting
$\left[\begin{array}{lll}
x=1 & ... & x=-3\\
y=(1)+2 & & y=(-3)+2\\
y=3 & & y=-1\\
& & \\
(1,3) & & (-3,-1)
\end{array}\right]$
Intersections: $(-3,-1)$ and $(1,3)$
