https://file.aiquickdraw.com/mathbot/file/1727777548912nt0ftxe1.jpg

The equation visible in the image is:

\[
3x^2 + 2y^2 - 6x - 12y + 15 = 0
\]

This appears to be the equation of a conic section. Let's analyze and simplify the equation to determine its type (ellipse, circle, hyperbola, or parabola) and potentially its center, axes, and orientation. 

To do this, we can attempt to complete the square for both \(x\) and \(y\).

### Steps to solve:
1. **Group \(x\)-terms** and **\(y\)-terms** together:
   \[
   (3x^2 - 6x) + (2y^2 - 12y) + 15 = 0
   \]

2. **Factor out constants** from the quadratic terms:
   \[
   3(x^2 - 2x) + 2(y^2 - 6y) + 15 = 0
   \]

3. **Complete the square** for \(x\)-terms and \(y\)-terms:
   - For \(x^2 - 2x\), add and subtract \((1)^2\):
     \[
     x^2 - 2x + 1 = (x - 1)^2 - 1
     \]
   - For \(y^2 - 6y\), add and subtract \((3)^2\):
     \[
     y^2 - 6y + 9 = (y - 3)^2 - 9
     \]

4. Substitute these completed squares back into the equation:
   \[
   3\left((x - 1)^2 - 1\right) + 2\left((y - 3)^2 - 9\right) + 15 = 0
   \]
   Simplify the equation:
   \[
   3(x - 1)^2 - 3 + 2(y - 3)^2 - 18 + 15 = 0
   \]
   \[
   3(x - 1)^2 + 2(y - 3)^2 - 6 = 0
   \]
   \[
   3(x - 1)^2 + 2(y - 3)^2 = 6
   \]
   \[
   \frac{(x - 1)^2}{2} + \frac{(y - 3)^2}{3} = 1
   \]

### Conclusion:
This is the equation of an **ellipse** with center at \((1, 3)\), semi-major axis \( \sqrt{3} \), and semi-minor axis \( \sqrt{2} \).

Would you like further details or have any questions on this problem?

### Relative Questions:
1. How do you determine the type of conic section from a general quadratic equation?
2. What are the steps to complete the square for a general quadratic equation?
3. How do you find the center and axes of an ellipse?
4. How do you derive the lengths of the semi-major and semi-minor axes from the equation of an ellipse?
5. What conditions make an ellipse become a circle?

### Tip:
When completing the square, always factor out coefficients from the squared terms before proceeding, as this simplifies the calculation and avoids errors.

Solve the equation 3x^2 + 2y^2 - 6x - 12y + 15 = 0

Learn how to solve the conic section equation 3x^2 + 2y^2 - 6x - 12y + 15 = 0 by completing the square. This solution helps you identify the equation as an ellipse and provides detailed steps to find the center and axes. Suitable for students in Grades 10-12.

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