SOLUTION: How do you find the point of intersection(s) for x = 2y^2 + 3y + 1 and 2x + 3y^2 = 0

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Question 1034344: How do you find the point of intersection(s) for x = 2y^2 + 3y + 1 and 2x + 3y^2 = 0
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
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.