SOLUTION: the sum twice the square of a positive odd integer and its consecutive is 57. what are the integers?
Algebra.Com
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) (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
= 21
x1=(-1+21)/4
x1=5
x2=(-1-21)/4
x2= -5.5
Ignore negative value
x = 5
5 &7
m.ananth@hotmail.ca
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