Combining the two terms on the left-hand side under a common denominator, we get \[\frac{x^2+11x+28 - 7(x+4)}{x+4} = x,\]or \[\frac{x^2+4x}{x+4} = x.\]If $ x \neq -4, $ then the left-hand side reduces to $ \frac{x(x+4)}{x+4} = x, $ so the equation is always true. And if $ x=-4, $ then the denominator of the left-hand side is zero, so the equation is not true. Hence, the solution set consists of all $ x $ such that $ x \neq -4 $. In interval notation, this is \[x \in \boxed{(-\infty, -4) \cup (-4, \infty)}.\]