Question 387221
{{{ x^2+y^2=4 }}} and {{{y=x^2-2}}}.
The 2nd equation becomes {{{y + 2=x^2}}}.  Substitute.
{{{y^2 + y + 2 = 4}}},  or {{{y^2 + y + -2 = 0}}}.
(y+2)(y - 1) = 0.
y = -2, y = 1.  When y = -2, then {{{-2 + 2 = x^2}}}, so x = 0.
When y = 1, then {{{1+2 = x^2}}}, or {{{x^2 = 3}}}, or x = {{{-sqrt(3)}}}, {{{sqrt(3)}}}.  Therefore there are 3 solutions: ({{{-sqrt(3)}}}, 1), ({{{sqrt(3)}}}, 1), and (0, -2).  The solutions correspond to the 3 points of intersection of the parabola {{{y=x^2-2}}} and the circle {{{ x^2+y^2=4 }}}.