Question 750196
{{{x^2=y+3}}}



{{{
2(y+3)-y^2=6
}}}




{{{y^2-2y=0}}}




so y=0 or  y=2




y=0:


{{{
x^2=y+3=3
}}}



{{{
x=sqrt(3)
}}}


or




{{{
x=-sqrt(3)
}}}



y=2:


{{{
x=sqrt(5)
}}}
or
{{{
x=-sqrt(5)
}}}


solutions


(-sqrt(3),0)
(sqrt(3),0)
(-sqrt(5),2)
(sqrt(5),2)




:)