Three brothers have ages that are consecutive intergers. 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.

1. write each boy's age in terms of x
Here's what we can do to represent each boy's age in terms of x:

1st brother = x + 2 (he is two years older than than 3nd brother)
2nd brother = x + 1 (he is one year older than 3rd brother)
3rd brother = x   (he is the youngest so we use x to represent him)

2. Write an equation to represent the situation:

1st brother's age times 3rd brother's age equals 20 more than twice 2nd brother's age

(1st brother's age)*(3rd brother's age) = 20 + 2(2nd brother's age) 

(x+2)*(x) = 20 + 2(x+1)

3. Now we need to solve for x.

(x+2)(x) = 20 + 2x +2
Foil the stuff in the parentheses

x^2 + 2x = 20 + 2x +2
Move everything to the same side

x^2 + 2x - 20 - 2x -2 = 20 + 2x +2  - 20 - 2x -2
all that crazy stuff becomes: x^2 - 22  = 0
Solve the polynomial

x^2 - 22 = 0
remember a^2 - b is the same as (a+√b)(a-√b) 
so x^2 - 22 = (x+√22)(x-√22)

This means (x+√22)(x-√22) = 0
so x = √22 or -√22

Age obviously won't be negative so x = √22 in this case. 
By the way: √22 also equals 4.69

4. Find out how old the three brothers are.

x = 3rd brother's age
x + 1 = 2nd brother's age
x + 2 = 1st brother's age
substitute √22 or 4.69 for x.

3rd brother = 4.69 years old (or √22 years old to be more precise)
2nd brother = 5.69 years old (or √22 + 1)
1st brother = 6.69 years old (or √22 +2)

Important NOTE: I think you made a mistake in typing the question. It might have been "consecutive even integers" rather than what you typed which was "consecutive integers". If the question contained "consecutive even integers"  than the brother's ages would have been 4, 6, and 8 instead of some funky decimal numbers.

Hope this answered your question, but if it was too confusing feel free to ask any questions you have.

Im pretty sure it’s is b 1.6 mi