1. **State the problem:** Find the equation of the hyperbola with foci at $$\left(-2 - \sqrt{45}, 3\right)$$ and $$\left(-2 + \sqrt{45}, 3\right)$$ and asymptotes $$y = 2x + 7$$ and $$y = -2x - 1$$. 2. **Identify the center:** The center $$ (h, k) $$ is the midpoint of the foci. $$h = \frac{-2 - \sqrt{45} + (-2 + \sqrt{45})}{2} = \frac{-4}{2} = -2$$ $$k = 3$$ So, center is $$(-2, 3)$$. 3. **Determine orientation:** Since foci share the same $$y$$-coordinate, the hyperbola opens horizontally. 4. **Recall hyperbola standard form (horizontal transverse axis):** $$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$$ 5. **Use asymptotes to find relationship between $$a$$ and $$b$$:** Asymptotes for horizontal hyperbola: $$y = k \pm \frac{b}{a}(x - h)$$ Given asymptotes: $$y = 2x + 7$$ and $$y = -2x - 1$$ Rewrite in form $$y - k = m(x - h)$$: $$y - 3 = 2(x + 2)$$ and $$y - 3 = -2(x + 2)$$ So slope $$m = \pm 2 = \pm \frac{b}{a}$$ Thus, $$\frac{b}{a} = 2 \implies b = 2a$$ 6. **Find focal distance $$c$$:** Distance between foci is $$2c$$. Given foci at $$x = -2 \pm \sqrt{45}$$, so $$c = \sqrt{45} = 3\sqrt{5}$$ 7. **Use relationship between $$a$$, $$b$$, and $$c$$:** $$c^2 = a^2 + b^2$$ Substitute $$b = 2a$$: $$c^2 = a^2 + (2a)^2 = a^2 + 4a^2 = 5a^2$$ 8. **Solve for $$a^2$$:** $$c^2 = 45 = 5a^2 \implies a^2 = 9$$ 9. **Find $$b^2$$:** $$b^2 = (2a)^2 = 4a^2 = 4 \times 9 = 36$$ 10. **Write the equation:** $$\frac{(x + 2)^2}{9} - \frac{(y - 3)^2}{36} = 1$$ **Final answer:** $$\boxed{\frac{(x + 2)^2}{9} - \frac{(y - 3)^2}{36} = 1}$$