The characteristic equation is \[\det\matrix{40-\lambda & -70 \\ 21 & -37-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 40-\lambda & -70 \\ 21 & -37-\lambda \end{array} \right\vert &= (40-\lambda)(-37-\lambda)+70\cdot21 \\
&= (40-\lambda)(-37-\lambda)+1470 \\ &= \lambda^2-3\,\lambda-10
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-3\,\lambda-10 = (\lambda-5)(\lambda+2) \] So, the eigenvalues are \(\lambda_1 = 5\) and \(\lambda_2 = -2\).

Let \(\lambda = 5\) be an eigenvalue of the matrix \[A=\matrix{40 & -70 \\ 21 & -37}\] 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 \[A-5 I = \matrix{40 & -70 \\ 21 & -37} - \matrix{5 & 0 \\0& 5 }=\matrix{ 35 & -70 \\ 21 & -42}\] This can be done as follows:
\[\begin{aligned}
\matrix{35&-70\\21&-42\\}&\sim\matrix{1&-2\\21&-42\\}&{\blue{\begin{array}{c}{{1}\over{35}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-2\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-21R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 5\) equals \(\left\{ r \cv{2\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{2\\1}\right\rangle\).

Let \(\lambda = -2\) be an eigenvalue of the matrix \[A=\matrix{40 & -70 \\ 21 & -37}\] 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 \[A+2 I = \matrix{40 & -70 \\ 21 & -37} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 42 & -70 \\ 21 & -35}\] This can be done as follows:
\[\begin{aligned}
\matrix{42&-70\\21&-35\\}&\sim\matrix{1&-{{5}\over{3}}\\21&-35\\}&{\blue{\begin{array}{c}{{1}\over{42}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{5}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-21R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -2\) equals \(\left\{ r \cv{5\\3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{5\\3}\right\rangle\).
If possible, we avoided fractions in the solution.