SOLUTION: I am having trouble figuring out how to work this problem out. Cannot get the right equations or answer. Find two consecutive even integers such that the square of the second,

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Question 864327: I am having trouble figuring out how to work this problem out. Cannot get the right equations or answer.
Find two consecutive even integers such that the square of the second, decreased by three times the first is 46.
Found 2 solutions by mananth, josgarithmetic:
Answer by mananth(16946) About Me  (Show Source):
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Find two consecutive even integers such that the square of the second, decreased by three times the first is 46.
let the integers be x & (x+2)
square of the second = (x+2) ^2
decreased by 3 times the first
(x+2)^2-3x46
x^2+4x+4-3x=46
x^2+x=46-4
x^2+x=42
x^2+x-42=0
x^2+7x-6x-42=0
x(x+7)-6(x+7)=0
(x+7)(x-6)=0
x=-7 OR x=6
even integers
so 6 & 8

Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
Integers, 2n and 2n+2.
%282n%2B2%29%5E2-3%2A2n, the square of the second decreased by three times the first;
%282n%2B2%29%5E2-6n=46, ... is 46.

4n%5E2%2B8n%2B4-6n=46
4n%5E2%2B2n%2B4-46=0
4n%5E2%2B2n-42=0
2n%5E2%2Bn-21=0
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%282n%2B7%29%28n-3%29=0
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n=3 making the two integers 6 and 8.