Question

the product of two consecutive even integers is 14 less than 7 times their sum. Find the 2 integers

Answer 1

if one is \(n\) the next is \(n+2\) and the product will be
\[n(n+2)\]

Answer 2

there sum is \(n+n+2=2n+2\) and so 7 times there sum is \(7(2n+2)\)
14 less than that amount is \(7(2n+2)-14\)

Answer 3

set
\[n(n+2)=7(2n+2)-14\] and solve for \(n\)

Answer 4

so...
n^2+2n=15n+14-14
n^2+2n=15n

Answer 5

n^2+2n-15n=0
n^2-13n=0