The characteristic equation is \[\det\matrix{-23-\lambda & 7 \\ -84 & 26-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -23-\lambda & 7 \\ -84 & 26-\lambda \end{array} \right\vert &= (-23-\lambda)(26-\lambda)-7\cdot-84 \\
&= (-23-\lambda)(26-\lambda)+588 \\ &= \lambda^2-3\,\lambda-10
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-3\,\lambda-10 = (\lambda+2)(\lambda-5) \] So, the eigenvalues are \(\lambda_1 = -2\) and \(\lambda_2 = 5\).

Let \(\lambda = -2\) be an eigenvalue of the matrix \[A=\matrix{-23 & 7 \\ -84 & 26}\] 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 \[A+2 I = \matrix{-23 & 7 \\ -84 & 26} - \matrix{-2 & 0 \\0& -2 }=\matrix{ -21 & 7 \\ -84 & 28}\] This can be done as follows:
\[\begin{aligned}
\matrix{-21&7\\-84&28\\}&\sim\matrix{1&-{{1}\over{3}}\\-84&28\\}&{\blue{\begin{array}{c}-{{1}\over{21}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{1}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+84R_1\end{array}}} \end{aligned}\] 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 \(\lambda = 5\) be an eigenvalue of the matrix \[A=\matrix{-23 & 7 \\ -84 & 26}\] 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 \[A-5 I = \matrix{-23 & 7 \\ -84 & 26} - \matrix{5 & 0 \\0& 5 }=\matrix{ -28 & 7 \\ -84 & 21}\] This can be done as follows:
\[\begin{aligned}
\matrix{-28&7\\-84&21\\}&\sim\matrix{1&-{{1}\over{4}}\\-84&21\\}&{\blue{\begin{array}{c}-{{1}\over{28}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{1}\over{4}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+84R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 5\) equals \(\left\{ r \cv{1\\4} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\4}\right\rangle\).
If possible, we avoided fractions in the solution.