Three consecutive odd intergers are such that the sum of the squares of the first two integers is 54 more than 20 times the third integer. Determine the three integers.

x^2 + (x+2)^2 = 54+20(x+4)

To solve this problem, let's break it down into steps.

Step 1: Represent the three consecutive odd integers
Let's assume the first odd integer is x. Since we are looking for consecutive odd integers, the next two odd integers would be x+2 and x+4.

Step 2: Set up the equation
According to the problem statement, the sum of the squares of the first two integers is 54 more than 20 times the third integer. Mathematically, we can represent this as:
x^2 + (x+2)^2 = 20(x+4) + 54

Step 3: Simplify and solve the equation
Expanding the squares and simplifying the equation, we get:
x^2 + x^2 + 4x + 4 = 20x + 80 + 54
2x^2 + 4x + 4 = 20x + 134
2x^2 - 16x - 130 = 0

Step 4: Solve the quadratic equation
To solve the quadratic equation, we can either factorize it or use the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)

Applying the values for a, b, and c from our equation, we get:
x = (-(-16) ± √((-16)^2 - 4*2*(-130))) / (2*2)
x = (16 ± √(256 + 1040)) / 4
x = (16 ± √1296) / 4
x = (16 ± 36) / 4

We get two possible values for x:
x1 = (16 + 36) / 4 = 52 / 4 = 13
x2 = (16 - 36) / 4 = -20 / 4 = -5

So, we have two possible sets of consecutive odd integers:
Set 1: 13, 15, 17
Set 2: -5, -3, -1