The characteristic equation is \[\det\matrix{43-\lambda & 120 \\ -15 & -42-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} 43-\lambda & 120 \\ -15 & -42-\lambda \end{array} \right\vert &= (43-\lambda)(-42-\lambda)-120\cdot-15 \\
&= (43-\lambda)(-42-\lambda)+1800 \\ &= \lambda^2-\lambda-6
\end{aligned}\] 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 \[A=\matrix{43 & 120 \\ -15 & -42}\] 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 \[A-3 I = \matrix{43 & 120 \\ -15 & -42} - \matrix{3 & 0 \\0& 3 }=\matrix{ 40 & 120 \\ -15 & -45}\] This can be done as follows:
\[\begin{aligned}
\matrix{40&120\\-15&-45\\}&\sim\matrix{1&3\\-15&-45\\}&{\blue{\begin{array}{c}{{1}\over{40}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&3\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+15R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{3\\-1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{3\\-1}\right\rangle\).

Let \(\lambda = -2\) be an eigenvalue of the matrix \[A=\matrix{43 & 120 \\ -15 & -42}\] 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 \[A+2 I = \matrix{43 & 120 \\ -15 & -42} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 45 & 120 \\ -15 & -40}\] This can be done as follows:
\[\begin{aligned}
\matrix{45&120\\-15&-40\\}&\sim\matrix{1&{{8}\over{3}}\\-15&-40\\}&{\blue{\begin{array}{c}{{1}\over{45}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{8}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+15R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -2\) equals \(\left\{ r \cv{-8\\3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-8\\3}\right\rangle\).
If possible, we avoided fractions in the solution.