To solve the quadratic equation \(5x^2 + 4xy + y^2 + 6x + 9 = 0\), we can approach it by completing the square or using other algebraic techniques. However, since it is a quadratic in two variables, it represents a conic section (ellipse, hyperbola, or parabola). Let's try to rewrite it in a more recognizable form.

First, let's group the terms involving \(x\):

\[5x^2 + 4xy + y^2 + 6x + 9 = 0\]

We can complete the square for the \(x\)-terms. To do this, we need to consider the terms \(5x^2 + 4xy + 6x\).

1. Factor out the coefficient of \(x^2\) from the \(x\)-terms:

\[5(x^2 + \frac{4}{5}xy) + y^2 + 6x + 9 = 0\]

2. To complete the square inside the parentheses, we need to add and subtract the square of half the coefficient of \(x\) in the term \(\frac{4}{5}xy\):

\[5\left(x^2 + \frac{4}{5}xy + \left(\frac{2}{5}y\right)^2 - \left(\frac{2}{5}y\right)^2\right) + y^2 + 6x + 9 = 0\]

\[5\left(\left(x + \frac{2}{5}y\right)^2 - \left(\frac{2}{5}y\right)^2\right) + y^2 + 6x + 9 = 0\]

3. Simplify the expression:

\[5\left(x + \frac{2}{5}y\right)^2 - 5\left(\frac{2}{5}y\right)^2 + y^2 + 6x + 9 = 0\]

\[5\left(x + \frac{2}{5}y\right)^2 - \frac{4}{5}y^2 + y^2 + 6x + 9 = 0\]

\[5\left(x + \frac{2}{5}y\right)^2 + \frac{1}{5}y^2 + 6x + 9 = 0\]

4. Combine the \(y\)-terms:

\[5\left(x + \frac{2}{5}y\right)^2 + \frac{6}{5}y^2 + 6x + 9 = 0\]

This form is still quite complex, and it might be more insightful to analyze the discriminant of the quadratic form to determine the type of conic section. The general quadratic form is \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\).

For our equation \(5x^2 + 4xy + y^2 + 6x + 9 = 0\):

- \(A = 5\)
- \(B = 4\)
- \(C = 1\)
- \(D = 6\)
- \(E = 0\)
- \(F = 9\)

The discriminant \(\Delta\) of the conic section is given by:

\[\Delta = B^2 - 4AC\]

Substitute the values:

\[\Delta = 4^2 - 4 \cdot 5 \cdot 1\]
\[\Delta = 16 - 20\]
\[\Delta = -4\]

Since \(\Delta < 0\), the conic section represented by the equation \(5x^2 + 4xy + y^2 + 6x + 9 = 0\) is an **ellipse**.

To find the exact form of the ellipse, further algebraic manipulation or transformation techniques would be required, but identifying the type of conic section is often a crucial first step.