Three brothers have ages that are consecutive even integers. the product of the first and third boys' ages is 20 more than wice the second boy's age. find the age of each of the three boys.

Let the first brother = x 
second brother = x + 2
third brother = x + 4

The product of the first and third boys' ages is 20 more than twice the second boy's age:

x(x+4) = 2(x+2) + 20
x² + 4x = 2x + 4 + 20
x² + 4x - 2x - 24 = 0
x² + 2x - 24 = 0
(x - 4)(x + 6) = 0
x = 4 or -6 (rejected, age cannot be negative)

First brother = 4
Second brother = 4 + 2 = 6
Third brother = 4 + 4 = 8

The three boys are 4, 6, and 8 years old. 

Three brothers have ages that are consecutive even integers. The product of the first and third boys' ages is 20 more than twice the second boy's age. Find the age of each of the three boys.
Let x be the age of the first boy, snce they are consecutive even integers, they will be 2 way from each other.
1st boy= x
2nd boy= x+2
3rd boy = x+4
"product means the answer to a multiplication problem;
x(x+4)=20+2(x+2)
x%5E2+4x=20+2x+4
x%5E2+4x-2x=20+4
x%5E2+2x=24
x%5E2+2x-24=0

(x+6)(x-4)=0
x+6=0
x=-6
x-4=0
x=4
SInce we know they are even integers, the answerhas to be;
x=4
so the first boys age = 4
2nd boy= 4+2=6
3rd boy= 4+4=8

Hope this helps you