The characteristic equation is \[\det\matrix{28-\lambda & 10 \\ -75 & -27-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 28-\lambda & 10 \\ -75 & -27-\lambda \end{array} \right\vert &= (28-\lambda)(-27-\lambda)-10\cdot-75 \\
&= (28-\lambda)(-27-\lambda)+750 \\ &= \lambda^2-\lambda-6
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-\lambda-6 = (\lambda+2)(\lambda-3) \] So, the eigenvalues are \(\lambda_1 = -2\) and \(\lambda_2 = 3\).

Let \(\lambda = -2\) be an eigenvalue of the matrix \[A=\matrix{28 & 10 \\ -75 & -27}\] 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 \[A+2 I = \matrix{28 & 10 \\ -75 & -27} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 30 & 10 \\ -75 & -25}\] This can be done as follows:
\[\begin{aligned}
\matrix{30&10\\-75&-25\\}&\sim\matrix{1&{{1}\over{3}}\\-75&-25\\}&{\blue{\begin{array}{c}{{1}\over{30}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+75R_1\end{array}}} \end{aligned}\] 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 \[A=\matrix{28 & 10 \\ -75 & -27}\] 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 \[A-3 I = \matrix{28 & 10 \\ -75 & -27} - \matrix{3 & 0 \\0& 3 }=\matrix{ 25 & 10 \\ -75 & -30}\] This can be done as follows:
\[\begin{aligned}
\matrix{25&10\\-75&-30\\}&\sim\matrix{1&{{2}\over{5}}\\-75&-30\\}&{\blue{\begin{array}{c}{{1}\over{25}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{2}\over{5}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+75R_1\end{array}}} \end{aligned}\] 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.