Find the range of values of $k$ for which $x^2 + 12x + 9$ is always greater than $4x + k$.

1. Set up the inequality and rearrange it into standard quadratic form

$x^2 + 12x + 9 > 4x + k$
$x^2 + 8x + (9 - k) > 0$

2. Apply the discriminant condition for a quadratic to be always positive

For the quadratic to be always positive, the discriminant must be less than zero ($b^2 - 4ac < 0$):
$8^2 - 4(1)(9 - k) < 0$

3. Solve the resulting linear inequality for $k$

$64 - 36 + 4k < 0$
$28 + 4k < 0$
$4k < -28$
$k < -7$