SOLUTION: Find two consecutive odd integers such that the square of the first added to 3 times the second is 24. n^2+3(n)=24 This is the equation I came up with

SOLUTION: Find two consecutive odd integers such that the square of the first added to 3 times the second is 24. n^2+3(n)=24 This is the equation I came up with

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Question 200271: Find two consecutive odd integers such that the square of the first added to 3 times the second is 24.
n^2+3(n)=24 This is the equation I came up with
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
n^2+3(n)=24
n^2+3n-24=0

x=(-3+-sqrt[3^2-4*1*-24])/2*1
x=(-3+-sqrt[9+96])/2
x=(-3+-sqrt105)/2
x=(-3+-10.247)/2
x=(-3+10.247)/2
x=7.247/2
x=3.6235 ans.
x=(-2-10.247)/2
x=-12.247/2
x=-6.1235 ans.
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