The characteristic equation is \[\det\matrix{8-\lambda & 2 \\ -12 & -2-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 8-\lambda & 2 \\ -12 & -2-\lambda \end{array} \right\vert &= (8-\lambda)(-2-\lambda)-2\cdot-12 \\
&= (8-\lambda)(-2-\lambda)+24 \\ &= \lambda^2-6\,\lambda+8
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-6\,\lambda+8 = (\lambda-2)(\lambda-4) \] So, the eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 4\).

Let \(\lambda = 2\) be an eigenvalue of the matrix \[A=\matrix{8 & 2 \\ -12 & -2}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=2\vec{v}\), this is, \[(A-2 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-2 I\). We can do this through row reduction of the matrix \[A-2 I = \matrix{8 & 2 \\ -12 & -2} - \matrix{2 & 0 \\0& 2 }=\matrix{ 6 & 2 \\ -12 & -4}\] This can be done as follows:
\[\begin{aligned}
\matrix{6&2\\-12&-4\\}&\sim\matrix{1&{{1}\over{3}}\\-12&-4\\}&{\blue{\begin{array}{c}{{1}\over{6}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 2\) equals \(\left\{ r \cv{1\\-3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\-3}\right\rangle\).

Let \(\lambda = 4\) be an eigenvalue of the matrix \[A=\matrix{8 & 2 \\ -12 & -2}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=4\vec{v}\), that is, \[(A-4 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-4 I\). We can do this through row reduction of the matrix \[A-4 I = \matrix{8 & 2 \\ -12 & -2} - \matrix{4 & 0 \\0& 4 }=\matrix{ 4 & 2 \\ -12 & -6}\] This can be done as follows:
\[\begin{aligned}
\matrix{4&2\\-12&-6\\}&\sim\matrix{1&{{1}\over{2}}\\-12&-6\\}&{\blue{\begin{array}{c}{{1}\over{4}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 4\) equals \(\left\{ r \cv{1\\-2} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\-2}\right\rangle\).
If possible, we avoided fractions in the solution.