Posted by Stupid on .
Find three consecutive even integers such that twice the product of the first and last numbers is 28 more than the square of the second number.
I know the answers i just need to know how to get to them and ive spent 30 min on it HELP ME!!!!

math 
bobpursley,
let the integers be n, n+1, and n+2
2n*(n+2)28=(n+1)^2
solve for n, notice it is a quadratic. 
math 
Stupid,
Buy why wouldn't it be n, n+2, and N+4 because it is even ingergers,

math 
Not answered,
this question is still not answered will some one walk through it step by step

math 
Reiny,
you 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 
math 
Adithya,
lets start with the equation
2n(n+4)  28 = (n+2)^2
using formula (a+b)^2=a^2+2ab+b^2
2n^2+8n28=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