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December 18, 2014

December 18, 2014

Posted by **Stupid** on Thursday, October 9, 2008 at 9:37pm.

I know the answers i just need to know how to get to them and ive spent 30 min on it HELP ME!!!!

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**bobpursley**, Thursday, October 9, 2008 at 9:43pmlet the integers be n, n+1, and n+2

2n*(n+2)-28=(n+1)^2

solve for n, notice it is a quadratic.

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**Stupid**, Thursday, October 9, 2008 at 9:45pmBuy why wouldn't it be n, n+2, and N+4 because it is even ingergers,

- math -
**Not answered**, Thursday, October 9, 2008 at 9:56pmthis question is still not answered will some one walk through it step by step

- math -
**Reiny**, Thursday, October 9, 2008 at 10:51pmyou are right, it should have been

n, n+2 and n+4 so let's adjust bobpursley's equation to look like

2n(n+4) - 28 = (n+2)^2

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- math -
**Adithya**, Friday, October 17, 2008 at 7:40pmlets start with the equation

2n(n+4) - 28 = (n+2)^2

using formula (a+b)^2=a^2+2ab+b^2

2n^2+8n-28=n^2+4n+32

simplyfying,

n2+4n=32

Solving for n, we get n=4

So the even consecutive numbers are 4,6,and 8.....

checking...

2(4*8)-28=36

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