anyone help? thankss

(x + 3)^2 <=1
x^2 + 6 x + 9<=1
-4<=x<=-2
x = -2, and x = -3 and x = - 4

If    (x + 3)²  ≤  1    then the absolute value of   x + 3   must be less than  1

| x + 3 |  ≤  1

Then in order for that to be true, it must be that

-1  ≤  x + 3  ≤  1

Subtract  3  from all parts of the inequality.

-4  ≤  x  ≤  -2

Then the integer solutions to that inequality are clearly

x = -4,  x = -3,  x = -2

Thanks guest and hectictar!!