Question 37227
Let call those integers x and y, if they are consecutives y=x+1.

x+(x+1)=x(x+1)-11 The sum of two consectutive positive integers is 11 less than their product.

2x+1=x^2+x-11
0=x^2-x-12

Using the general formula for solving quadratic equations we get x1=4,x2=-3

Then if the integers are positive they are 4 and 5, 4+5=9, 4*5=20, 20-9=11,
And if they are negative they are -3 and -2, -3+(-2)=-5, -3*(-2)=6, 6-(-5)=11.