SOLUTION: the sum twice the square of a positive odd integer and its consecutive is 57. what are the integers?

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Question 447414: the sum twice the square of a positive odd integer and its consecutive is 57. what are the integers?
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
2x^2+x+2=57
2x^2+x-55=0
a= 2 b= 1 c= -55

b^2-4ac= 1+440
b^2-4ac= 441
sqrt%28%09441%29= 21
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=(-1+21)/4
x1=5
x2=(-1-21)/4
x2= -5.5
Ignore negative value
x = 5
5 &7
m.ananth@hotmail.ca