SOLUTION: Find the points of intersection between Y= 2x^2 + 3x + 4 and Y= x^2 + 6x + 2

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Question 640504: Find the points of intersection between Y= 2x^2 + 3x + 4 and Y= x^2 + 6x + 2
Answer by lwsshak3(11628) About Me  (Show Source):
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Find the points of intersection between Y= 2x^2 + 3x + 4 and Y= x^2 + 6x + 2
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Y= 2x^2 + 3x + 4
Y= x^2 + 6x + 2
..
equate right sides of both equations
2x^2 + 3x + 4=x^2 + 6x + 2
x^2-3x+2=0
(x-2)(x-1)=0
..
For x=2
y= x^2+6x+2=4+12+2=18
..
For x=1
y= x^2+6x+2=1+6+2=9
..
2 points of intersection: (2,18) and (1,9)