The characteristic equation is \[\det\matrix{22-\lambda & 24 \\ -18 & -20-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 22-\lambda & 24 \\ -18 & -20-\lambda \end{array} \right\vert &= (22-\lambda)(-20-\lambda)-24\cdot-18 \\
&= (22-\lambda)(-20-\lambda)+432 \\ &= \lambda^2-2\,\lambda-8
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-2\,\lambda-8 = (\lambda+2)(\lambda-4) \] So, the eigenvalues are \(\lambda_1 = -2\) and \(\lambda_2 = 4\).

Let \(\lambda = -2\) be an eigenvalue of the matrix \[A=\matrix{22 & 24 \\ -18 & -20}\] 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 \[A+2 I = \matrix{22 & 24 \\ -18 & -20} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 24 & 24 \\ -18 & -18}\] This can be done as follows:
\[\begin{aligned}
\matrix{24&24\\-18&-18\\}&\sim\matrix{1&1\\-18&-18\\}&{\blue{\begin{array}{c}{{1}\over{24}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+18R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -2\) equals \(\left\{ r \cv{1\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\-1}\right\rangle\).

Let \(\lambda = 4\) be an eigenvalue of the matrix \[A=\matrix{22 & 24 \\ -18 & -20}\] 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 \[A-4 I = \matrix{22 & 24 \\ -18 & -20} - \matrix{4 & 0 \\0& 4 }=\matrix{ 18 & 24 \\ -18 & -24}\] This can be done as follows:
\[\begin{aligned}
\matrix{18&24\\-18&-24\\}&\sim\matrix{1&{{4}\over{3}}\\-18&-24\\}&{\blue{\begin{array}{c}{{1}\over{18}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{4}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+18R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 4\) equals \(\left\{ r \cv{4\\-3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{4\\-3}\right\rangle\).
If possible, we avoided fractions in the solution.