Question 1034344
Well one way is to substitute
x = 2y^2 + 3y + 1 into 2x + 3y^2 = 0 and get
2(2y^2 + 3y + 1) + 3y^2 = 0
7y^2 + 6y + 2 = 0
Since y has no real roots, I'm thinking these graphs do not intersect at all.