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

Algebra ->  Equations -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
n^2+3(n)=24
n^2+3n-24=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
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.