The characteristic equation is \[\det\matrix{3-\lambda & 2 \\ -21 & -10-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 3-\lambda & 2 \\ -21 & -10-\lambda \end{array} \right\vert &= (3-\lambda)(-10-\lambda)-2\cdot-21 \\
&= (3-\lambda)(-10-\lambda)+42 \\ &= \lambda^2+7\,\lambda+12
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2+7\,\lambda+12 = (\lambda+3)(\lambda+4) \] So, the eigenvalues are \(\lambda_1 = -3\) and \(\lambda_2 = -4\).

Let \(\lambda = -3\) be an eigenvalue of the matrix \[A=\matrix{3 & 2 \\ -21 & -10}\] 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 \[A+3 I = \matrix{3 & 2 \\ -21 & -10} - \matrix{-3 & 0 \\0& -3 }=\matrix{ 6 & 2 \\ -21 & -7}\] This can be done as follows:
\[\begin{aligned}
\matrix{6&2\\-21&-7\\}&\sim\matrix{1&{{1}\over{3}}\\-21&-7\\}&{\blue{\begin{array}{c}{{1}\over{6}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+21R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -3\) equals \(\left\{ r \cv{1\\-3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{1\\-3}\right\rangle\).

Let \(\lambda = -4\) be an eigenvalue of the matrix \[A=\matrix{3 & 2 \\ -21 & -10}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-4\vec{v}\), that is, \[(A+4 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+4 I\). We can do this through row reduction of the matrix \[A+4 I = \matrix{3 & 2 \\ -21 & -10} - \matrix{-4 & 0 \\0& -4 }=\matrix{ 7 & 2 \\ -21 & -6}\] This can be done as follows:
\[\begin{aligned}
\matrix{7&2\\-21&-6\\}&\sim\matrix{1&{{2}\over{7}}\\-21&-6\\}&{\blue{\begin{array}{c}{{1}\over{7}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{2}\over{7}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+21R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -4\) equals \(\left\{ r \cv{-2\\7} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-2\\7}\right\rangle\).
If possible, we avoided fractions in the solution.