The characteristic equation is \[\det\matrix{3-\lambda & 5 \\ 0 & -2-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 3-\lambda & 5 \\ 0 & -2-\lambda \end{array} \right\vert &= (3-\lambda)(-2-\lambda)-5\cdot0 \\
&= (3-\lambda)(-2-\lambda)
\end{aligned}\] So, the eigenvalues are \(\lambda_1 = -2\) and \(\lambda_2 = 3\).

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

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