Three consecutive odd integers are such that the square of the third integer is 1515 greater thangreater than the sum of the squares of the first two. one solution is 11, 33, and 55. find three other consecutive odd integers that also satisfy the given conditions.

3 consecutive odd integersx, x + 2, and x + 4

(x + 4)^2 = (x^2 + (x + 2)^2) + 15
x^2 + 8x + 16 = x^2 + x^2 + 4x + 4 + 15
x^2 + 8x + 16 = 2x^2 + 4x + 19
2x^2 + 4x + 19 - x^2 - 8x - 16 = 0
x^2 - 4x + 3 = 0
(x - 3)(x - 1) = 0

x - 3 = 0x + 2 = 3 + 2 = 5x + 4 = 3 + 4 = 7
x = 3
so another one that satisfies this is : 3,5,7 <===

x - 1 = 0      x + 2 = 1 + 2 = 3x + 4 = 1 + 4 = 5
x = 1
and another one is : 1,3,5 <===

The probability of picking 2 green markers is 2/10