1. **State the problem:** Find the center and vertices of the ellipse given by the equation $$3x^2 + 2y^2 - 24x + 8y + 20 = 0$$. 2. **Rewrite the equation grouping x and y terms:** $$3x^2 - 24x + 2y^2 + 8y + 20 = 0$$ 3. **Divide the entire equation by the coefficients of the squared terms to complete the square:** $$3(x^2 - 8x) + 2(y^2 + 4y) + 20 = 0$$ 4. **Complete the square for x and y terms:** - For $x^2 - 8x$, half of $-8$ is $-4$, square it to get $16$. - For $y^2 + 4y$, half of $4$ is $2$, square it to get $4$. Add and subtract these inside the equation, remembering to balance the equation: $$3(x^2 - 8x + 16 - 16) + 2(y^2 + 4y + 4 - 4) + 20 = 0$$ 5. **Rewrite as:** $$3((x - 4)^2 - 16) + 2((y + 2)^2 - 4) + 20 = 0$$ 6. **Distribute and simplify:** $$3(x - 4)^2 - 48 + 2(y + 2)^2 - 8 + 20 = 0$$ $$3(x - 4)^2 + 2(y + 2)^2 - 36 = 0$$ 7. **Add 36 to both sides:** $$3(x - 4)^2 + 2(y + 2)^2 = 36$$ 8. **Divide both sides by 36:** $$\frac{3(x - 4)^2}{36} + \frac{2(y + 2)^2}{36} = \frac{36}{36}$$ Simplify fractions: $$\frac{(x - 4)^2}{12} + \frac{(y + 2)^2}{18} = 1$$ 9. **Identify the center:** The ellipse is centered at $$\boxed{(4, -2)}$$. 10. **Identify $a^2$ and $b^2$:** $$a^2 = 18, \quad b^2 = 12$$ Since $a^2 > b^2$, the major axis is along the y-direction. 11. **Calculate the vertices:** - Vertices along y-axis: $$y = -2 \pm \sqrt{18} = -2 \pm 3\sqrt{2}$$ - Vertices coordinates: $$\boxed{(4, -2 + 3\sqrt{2}) \text{ and } (4, -2 - 3\sqrt{2})}$$ **Final answer:** - Center: $(4, -2)$ - Vertices: $(4, -2 + 3\sqrt{2})$ and $(4, -2 - 3\sqrt{2})$