SOLUTION: please help me solve this problem: two consecutive positive integers have the property that one integer times twice the other equals 612. What is the sum of these two integers?

Question 188491: please help me solve this problem:
two consecutive positive integers have the property that one integer times twice the other equals 612. What is the sum of these two integers?

thank you!
Found 2 solutions by Alan3354, Mathtut:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
call the 1st integer n (i is too easily confused, imo)
The 2nd integer is n+1.
Twice this integer is 2n+2
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n*(2n+2) = 612
2n^2 + 2n = 612
n^2 + n = 306
n^2 + n - 306 = 0
(n+18)*(n-17) = 0
n = -18 (we're looking for positive integers, so ignore -18)
n = 17 That one works.
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The 2 integers are 17 & 18, so the sum is 35

Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
lets call our intergers a and a+1
:
a*2(a+1)=612
:
a(2a+2)=612
:

:

:

:
so a = -18 and 17......we are looking for postive interger
:
so a=17
and a+1=18
:
so our two integers are 17 and 18
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