1. **State the problem:** Find the points of intersection between the circle $$(x+2)^2 + (y-4)^2 = 5$$ and the line $$y = -2x + 4$$. 2. **Formula and rules:** To find intersection points, substitute the expression for $$y$$ from the line equation into the circle equation and solve for $$x$$. 3. **Substitute:** Replace $$y$$ in the circle equation: $$ (x+2)^2 + ((-2x + 4) - 4)^2 = 5 $$ 4. **Simplify inside the second squared term:** $$ (x+2)^2 + (-2x)^2 = 5 $$ 5. **Expand the squares:** $$ (x+2)^2 = x^2 + 4x + 4 $$ $$ (-2x)^2 = 4x^2 $$ 6. **Write the equation:** $$ x^2 + 4x + 4 + 4x^2 = 5 $$ 7. **Combine like terms:** $$ 5x^2 + 4x + 4 = 5 $$ 8. **Bring all terms to one side:** $$ 5x^2 + 4x + 4 - 5 = 0 $$ $$ 5x^2 + 4x - 1 = 0 $$ 9. **Solve quadratic equation using the quadratic formula:** $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $$a=5$$, $$b=4$$, $$c=-1$$. 10. **Calculate discriminant:** $$ \Delta = 4^2 - 4 \times 5 \times (-1) = 16 + 20 = 36 $$ 11. **Calculate roots:** $$ x = \frac{-4 \pm \sqrt{36}}{2 \times 5} = \frac{-4 \pm 6}{10} $$ 12. **Find each root:** - $$ x_1 = \frac{-4 + 6}{10} = \frac{2}{10} = 0.2 $$ - $$ x_2 = \frac{-4 - 6}{10} = \frac{-10}{10} = -1 $$ 13. **Find corresponding $$y$$ values using $$y = -2x + 4$$:** - For $$x_1 = 0.2$$: $$ y_1 = -2(0.2) + 4 = -0.4 + 4 = 3.6 $$ - For $$x_2 = -1$$: $$ y_2 = -2(-1) + 4 = 2 + 4 = 6 $$ 14. **Final answer:** The points of intersection are $$\boxed{(0.2, 3.6)}$$ and $$\boxed{(-1, 6)}$$.