1. **State the problem:** Find the eigenvalues $\lambda_1$ and $\lambda_2$ of the matrix $$\begin{bmatrix} 3 & 3 \\ 3 & 4 \end{bmatrix}.$$\n\n2. **Formula:** Eigenvalues satisfy the characteristic equation $$\det(A - \lambda I) = 0,$$ where $A$ is the matrix and $I$ is the identity matrix.\n\n3. **Set up the characteristic polynomial:**\n$$\det\left(\begin{bmatrix} 3 & 3 \\ 3 & 4 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) = \det\begin{bmatrix} 3 - \lambda & 3 \\ 3 & 4 - \lambda \end{bmatrix} = 0.$$\n\n4. **Calculate the determinant:**\n$$ (3 - \lambda)(4 - \lambda) - 3 \times 3 = 0.$$\n\n5. **Expand:**\n$$ 12 - 3\lambda - 4\lambda + \lambda^2 - 9 = 0,$$\nwhich simplifies to\n$$ \lambda^2 - 7\lambda + 3 = 0.$$\n\n6. **Solve the quadratic equation:**\nUsing the quadratic formula $$\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $a=1$, $b=-7$, $c=3$, we get\n$$\lambda = \frac{7 \pm \sqrt{49 - 12}}{2} = \frac{7 \pm \sqrt{37}}{2}.$$\n\n7. **Final eigenvalues:**\n$$\lambda_1 = \frac{7 + \sqrt{37}}{2}, \quad \lambda_2 = \frac{7 - \sqrt{37}}{2}.$$