What is two consecutive integers such that the square of the larger integer is 19 more than 9 times the smaller integer?

Answer 1

9 and 10

let the 1st integer be = a

Let the 2nd integer be = a+1

(a+1)² - 9a= 19

a² +1 +2a-9a=19

a ² - 9a + 2a -18=0

a(a-9)+2 (a-9)=0

(a-9)(a+2)=0

Since an integer is a positive number a=9 and a+1=10.