The characteristic equation is \[\det\matrix{-8-\lambda & 2 \\ -15 & 3-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -8-\lambda & 2 \\ -15 & 3-\lambda \end{array} \right\vert &= (-8-\lambda)(3-\lambda)-2\cdot-15 \\
&= (-8-\lambda)(3-\lambda)+30 \\ &= \lambda^2+5\,\lambda+6
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2+5\,\lambda+6 = (\lambda+2)(\lambda+3) \] So, the eigenvalues are \(\lambda_1 = -2\) and \(\lambda_2 = -3\).

Let \(\lambda = -2\) be an eigenvalue of the matrix \[A=\matrix{-8 & 2 \\ -15 & 3}\] 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 \[A+2 I = \matrix{-8 & 2 \\ -15 & 3} - \matrix{-2 & 0 \\0& -2 }=\matrix{ -6 & 2 \\ -15 & 5}\] This can be done as follows:
\[\begin{aligned}
\matrix{-6&2\\-15&5\\}&\sim\matrix{1&-{{1}\over{3}}\\-15&5\\}&{\blue{\begin{array}{c}-{{1}\over{6}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{1}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+15R_1\end{array}}} \end{aligned}\] 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 \(\lambda = -3\) be an eigenvalue of the matrix \[A=\matrix{-8 & 2 \\ -15 & 3}\] 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 \[A+3 I = \matrix{-8 & 2 \\ -15 & 3} - \matrix{-3 & 0 \\0& -3 }=\matrix{ -5 & 2 \\ -15 & 6}\] This can be done as follows:
\[\begin{aligned}
\matrix{-5&2\\-15&6\\}&\sim\matrix{1&-{{2}\over{5}}\\-15&6\\}&{\blue{\begin{array}{c}-{{1}\over{5}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{2}\over{5}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+15R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -3\) equals \(\left\{ r \cv{-2\\-5} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-2\\-5}\right\rangle\).
If possible, we avoided fractions in the solution.