Find two consecutive positive integers such that the square of that the square of the larger integers is nineteen more than nine times the smaller integer.

Let the consecutive positive integers be x and x+1.
According to the question,
(x+1)^2=19+9x
x^2+2x+1=19+9x
x^2-7x-18=0
(x-9) and (x+2)=0
x=9 and x=-2(rejected)
Therefore, the consecutive positive integers are 9 and 10.

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