The characteristic equation is \[\det\matrix{18-\lambda & 15 \\ -20 & -17-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 18-\lambda & 15 \\ -20 & -17-\lambda \end{array} \right\vert &= (18-\lambda)(-17-\lambda)-15\cdot-20 \\
&= (18-\lambda)(-17-\lambda)+300 \\ &= \lambda^2-\lambda-6
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-\lambda-6 = (\lambda-3)(\lambda+2) \] So, the eigenvalues are \(\lambda_1 = 3\) and \(\lambda_2 = -2\).

Let \(\lambda = 3\) be an eigenvalue of the matrix \[A=\matrix{18 & 15 \\ -20 & -17}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=3\vec{v}\), this 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 \[A-3 I = \matrix{18 & 15 \\ -20 & -17} - \matrix{3 & 0 \\0& 3 }=\matrix{ 15 & 15 \\ -20 & -20}\] This can be done as follows:
\[\begin{aligned}
\matrix{15&15\\-20&-20\\}&\sim\matrix{1&1\\-20&-20\\}&{\blue{\begin{array}{c}{{1}\over{15}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&1\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+20R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{1\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\-1}\right\rangle\).

Let \(\lambda = -2\) be an eigenvalue of the matrix \[A=\matrix{18 & 15 \\ -20 & -17}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-2\vec{v}\), that 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 \[A+2 I = \matrix{18 & 15 \\ -20 & -17} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 20 & 15 \\ -20 & -15}\] This can be done as follows:
\[\begin{aligned}
\matrix{20&15\\-20&-15\\}&\sim\matrix{1&{{3}\over{4}}\\-20&-15\\}&{\blue{\begin{array}{c}{{1}\over{20}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{3}\over{4}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+20R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -2\) equals \(\left\{ r \cv{-3\\4} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-3\\4}\right\rangle\).
If possible, we avoided fractions in the solution.