The characteristic equation is $$\det\matrix{-35-\lambda & -12 \\ 96 & 33-\lambda }=0$$ We first rewrite the characteristic polynomial of $A$:
$$\begin{aligned} \det(A-\lambda I) = \left\vert \begin{array}{cc} -35-\lambda & -12 \\ 96 & 33-\lambda \end{array} \right\vert &= (-35-\lambda)(33-\lambda)+12\cdot96 \\ &= (-35-\lambda)(33-\lambda)+1152 \\ &= \lambda^2+2\,\lambda-3 \end{aligned}$$ To solve this quadratic equation, we can factor it in the following way: $$\lambda^2+2\,\lambda-3 = (\lambda-1)(\lambda+3)$$ So, the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = -3$.

Let $\lambda = 1$ be an eigenvalue of the matrix $$A=\matrix{-35 & -12 \\ 96 & 33}$$ Then there must be a vector $\vec{v}$ such that $A\vec{v}=1\vec{v}$, this is, $$(A-I)\vec{v}=\vec{0}\text.$$ In other words, we must find the kernel of the matrix $A-I$. We can do this through row reduction of the matrix $$A-I = \matrix{-35 & -12 \\ 96 & 33} - \matrix{1 & 0 \\0& 1 }=\matrix{ -36 & -12 \\ 96 & 32}$$ This can be done as follows:
$$\begin{aligned} \matrix{-36&-12\\96&32\\}&\sim\matrix{1&{{1}\over{3}}\\96&32\\}&{\blue{\begin{array}{c}-{{1}\over{36}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-96R_1\end{array}}} \end{aligned}$$ So the eigenspace for $\lambda = 1$ 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 $$A=\matrix{-35 & -12 \\ 96 & 33}$$ 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 $$A+3 I = \matrix{-35 & -12 \\ 96 & 33} - \matrix{-3 & 0 \\0& -3 }=\matrix{ -32 & -12 \\ 96 & 36}$$ This can be done as follows:
$$\begin{aligned} \matrix{-32&-12\\96&36\\}&\sim\matrix{1&{{3}\over{8}}\\96&36\\}&{\blue{\begin{array}{c}-{{1}\over{32}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{3}\over{8}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-96R_1\end{array}}} \end{aligned}$$ So the eigenspace for $\lambda = -3$ equals $\left\{ r \cv{3\\-8} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{3\\-8}\right\rangle$.
If possible, we avoided fractions in the solution.