The characteristic equation is \[\det\matrix{-7-\lambda & -6 \\ 20 & 15-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -7-\lambda & -6 \\ 20 & 15-\lambda \end{array} \right\vert &= (-7-\lambda)(15-\lambda)+6\cdot20 \\
&= (-7-\lambda)(15-\lambda)+120 \\ &= \lambda^2-8\,\lambda+15
\end{aligned}\] 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 \[A=\matrix{-7 & -6 \\ 20 & 15}\] 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 \[A-5 I = \matrix{-7 & -6 \\ 20 & 15} - \matrix{5 & 0 \\0& 5 }=\matrix{ -12 & -6 \\ 20 & 10}\] This can be done as follows:
\[\begin{aligned}
\matrix{-12&-6\\20&10\\}&\sim\matrix{1&{{1}\over{2}}\\20&10\\}&{\blue{\begin{array}{c}-{{1}\over{12}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-20R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 5\) equals \(\left\{ r \cv{1\\-2} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\-2}\right\rangle\).

Let \(\lambda = 3\) be an eigenvalue of the matrix \[A=\matrix{-7 & -6 \\ 20 & 15}\] 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 \[A-3 I = \matrix{-7 & -6 \\ 20 & 15} - \matrix{3 & 0 \\0& 3 }=\matrix{ -10 & -6 \\ 20 & 12}\] This can be done as follows:
\[\begin{aligned}
\matrix{-10&-6\\20&12\\}&\sim\matrix{1&{{3}\over{5}}\\20&12\\}&{\blue{\begin{array}{c}-{{1}\over{10}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{3}\over{5}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-20R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{3\\-5} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{3\\-5}\right\rangle\).
If possible, we avoided fractions in the solution.