The characteristic equation is \[\det\matrix{-9-\lambda & -12 \\ 2 & 1-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -9-\lambda & -12 \\ 2 & 1-\lambda \end{array} \right\vert &= (-9-\lambda)(1-\lambda)+12\cdot2 \\
&= (-9-\lambda)(1-\lambda)+24 \\ &= \lambda^2+8\,\lambda+15
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2+8\,\lambda+15 = (\lambda+5)(\lambda+3) \] So, the eigenvalues are \(\lambda_1 = -5\) and \(\lambda_2 = -3\).

Let \(\lambda = -5\) be an eigenvalue of the matrix \[A=\matrix{-9 & -12 \\ 2 & 1}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-5\vec{v}\), this is, \[(A+5 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+5 I\). We can do this through row reduction of the matrix \[A+5 I = \matrix{-9 & -12 \\ 2 & 1} - \matrix{-5 & 0 \\0& -5 }=\matrix{ -4 & -12 \\ 2 & 6}\] This can be done as follows:
\[\begin{aligned}
\matrix{-4&-12\\2&6\\}&\sim\matrix{1&3\\2&6\\}&{\blue{\begin{array}{c}-{{1}\over{4}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&3\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-2R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -5\) equals \(\left\{ r \cv{3\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{3\\-1}\right\rangle\).

Let \(\lambda = -3\) be an eigenvalue of the matrix \[A=\matrix{-9 & -12 \\ 2 & 1}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-3\vec{v}\), that is, \[(A+3 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+3 I\). We can do this through row reduction of the matrix \[A+3 I = \matrix{-9 & -12 \\ 2 & 1} - \matrix{-3 & 0 \\0& -3 }=\matrix{ -6 & -12 \\ 2 & 4}\] This can be done as follows:
\[\begin{aligned}
\matrix{-6&-12\\2&4\\}&\sim\matrix{1&2\\2&4\\}&{\blue{\begin{array}{c}-{{1}\over{6}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&2\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-2R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -3\) equals \(\left\{ r \cv{-2\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-2\\1}\right\rangle\).
If possible, we avoided fractions in the solution.