The characteristic equation is \[\det\matrix{-27-\lambda & 24 \\ -32 & 29-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -27-\lambda & 24 \\ -32 & 29-\lambda \end{array} \right\vert &= (-27-\lambda)(29-\lambda)-24\cdot-32 \\
&= (-27-\lambda)(29-\lambda)+768 \\ &= \lambda^2-2\,\lambda-15
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-2\,\lambda-15 = (\lambda+3)(\lambda-5) \] So, the eigenvalues are \(\lambda_1 = -3\) and \(\lambda_2 = 5\).

Let \(\lambda = -3\) be an eigenvalue of the matrix \[A=\matrix{-27 & 24 \\ -32 & 29}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-3\vec{v}\), this 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 \[A+3 I = \matrix{-27 & 24 \\ -32 & 29} - \matrix{-3 & 0 \\0& -3 }=\matrix{ -24 & 24 \\ -32 & 32}\] This can be done as follows:
\[\begin{aligned}
\matrix{-24&24\\-32&32\\}&\sim\matrix{1&-1\\-32&32\\}&{\blue{\begin{array}{c}-{{1}\over{24}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+32R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -3\) equals \(\left\{ r \cv{1\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\1}\right\rangle\).

Let \(\lambda = 5\) be an eigenvalue of the matrix \[A=\matrix{-27 & 24 \\ -32 & 29}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=5\vec{v}\), that 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 \[A-5 I = \matrix{-27 & 24 \\ -32 & 29} - \matrix{5 & 0 \\0& 5 }=\matrix{ -32 & 24 \\ -32 & 24}\] This can be done as follows:
\[\begin{aligned}
\matrix{-32&24\\-32&24\\}&\sim\matrix{1&-{{3}\over{4}}\\-32&24\\}&{\blue{\begin{array}{c}-{{1}\over{32}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{3}\over{4}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+32R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 5\) equals \(\left\{ r \cv{3\\4} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{3\\4}\right\rangle\).
If possible, we avoided fractions in the solution.