SOLUTION: find two consecutive positive odd numbers which are such that the square of their sum exceeds the sum of their squares by 126

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Question 449169: find two consecutive positive odd numbers which are such that the square of their sum exceeds the sum of their squares by 126
Answer by mananth(16946) About Me  (Show Source):
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x
x+2
...
(x+x+2)^2=x^2+(x+2)^2+126
(2x+2)^2=x^2+x^2+4x+4+126
4x^2+8x+4=2x^2+4x+130
2x^2+4x-126=0
a=2,b=4,c=-126

b^2-4ac=16+1008
b^2-4ac=1024
sqrt( 1024 )= 32
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=(-4+ 32)/4
x1=7
x2=(-4-32)/4
x2= -9
7 & 9