The characteristic equation is \[\det\matrix{-15-\lambda & 30 \\ -4 & 7-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -15-\lambda & 30 \\ -4 & 7-\lambda \end{array} \right\vert &= (-15-\lambda)(7-\lambda)-30\cdot-4 \\
&= (-15-\lambda)(7-\lambda)+120 \\ &= \lambda^2+8\,\lambda+15
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2+8\,\lambda+15 = (\lambda+5)(\lambda+3) \] So, the eigenvalues are \(\lambda_1 = -5\) and \(\lambda_2 = -3\).

Let \(\lambda = -5\) be an eigenvalue of the matrix \[A=\matrix{-15 & 30 \\ -4 & 7}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-5\vec{v}\), this 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 \[A+5 I = \matrix{-15 & 30 \\ -4 & 7} - \matrix{-5 & 0 \\0& -5 }=\matrix{ -10 & 30 \\ -4 & 12}\] This can be done as follows:
\[\begin{aligned}
\matrix{-10&30\\-4&12\\}&\sim\matrix{1&-3\\-4&12\\}&{\blue{\begin{array}{c}-{{1}\over{10}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-3\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+4R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -5\) equals \(\left\{ r \cv{-3\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-3\\-1}\right\rangle\).

Let \(\lambda = -3\) be an eigenvalue of the matrix \[A=\matrix{-15 & 30 \\ -4 & 7}\] 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 \[A+3 I = \matrix{-15 & 30 \\ -4 & 7} - \matrix{-3 & 0 \\0& -3 }=\matrix{ -12 & 30 \\ -4 & 10}\] This can be done as follows:
\[\begin{aligned}
\matrix{-12&30\\-4&10\\}&\sim\matrix{1&-{{5}\over{2}}\\-4&10\\}&{\blue{\begin{array}{c}-{{1}\over{12}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{5}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+4R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -3\) equals \(\left\{ r \cv{-5\\-2} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-5\\-2}\right\rangle\).
If possible, we avoided fractions in the solution.