Question 579722
let the integers be x & (x+2)

x(x+2)+2=290

x^2+2x+2=290

Add -2 to both sides

x^2+2x-288=0

x^2+18x-16x-288=0

x(x+18)-16(x+18)=0
(x+18)(x-16)=0

x=-18, x=16

The numbers are 16 & 18

OR
-18, -16