Question 820501
Three brothers have ages that are consecutive even integers.
x, (x+2), (x+4) are the ages of the 3 brothers
:
 The product of the first and third boy's ages is 20 more than twice the second boys age, 
x(x+4) = 2(x+2) + 20
x^2 + 4x = 2x + 4 + 20
x^2 + 4x = 2x + 24
Combine as a quadratic equation on the left
x^2 + 4x - 2x - 24 = 0
x^2 + 2x = 24 = 0
Factors to
(x+6)(x-4) = 0
The positive solution
x = 4 is the youngest, I'll let you find the ages of the other two
Check your answers in the given statement