Chapter 9 - Analytic Geometry - Section 9.2 The Parabola - 9.2 Assess Your Understanding - Page 667: 3

{$−7,−1$}

Take the square root of both sides of the equation $(x+4)^2=9$. The real solutions of the equation $(x+4)^2=9$ can be found with the Square Root Method. $(x+4)^2=\sqrt 9 \\ x+4 = \pm 3 $ Now, solve for $x$ . $x=-4 + 3, -4 -3 \\ x=-1, -7$ Therefore, the required solution set is {$−7,−1$}