Finding the eigenvalues and a basis for the eigenspaces of a $3\times3$ matrix. For the  matrix $A \in M_{3\times3}(\mathbb{R})$ below, I need to find the eigenvalues and a basis for the corresponding eigenspaces:
$$\begin{bmatrix}\
1 & -3 & 3 \\           
3 & -5 & 3 \\
6 & -6 & 4 \\
\end{bmatrix}$$
I have tried to use the formula $\det(I\lambda - A) = 0$ but I ended up with equation $\lambda^3 - 12\lambda - 16 = 0$, of which I can't seem to find the solutions to.

You can use the rationnal root theorem to find a root of your polynomial : $x$ is a rationnal root of $\lambda^3 - 12\lambda - 16\Rightarrow  x$ is a divisor of $16$. You check if there is a root of $\lambda^3 - 12\lambda - 16$ in $\{\pm1,\pm2,\pm4,\pm8,\pm16\}$, and you find that $4$ is indeed a root. Thus $\lambda^3 - 12\lambda - 16=(\lambda-4)(\lambda^2+4\lambda+4)$. 
Can you finish ?
The eigenvalues are $-2$ and $4$

Now will find the eigen sace associated to $4$ for example. $X=\begin{pmatrix} x \\ y\\z \end{pmatrix}$ is in this eigenspace iff :
$$AX=4X\iff \begin{pmatrix}\
1 & -3 & 3 \\           
3 & -5 & 3 \\
6 & -6 & 4 \\
\end{pmatrix}\begin{pmatrix} x \\ y\\z \end{pmatrix}=4\begin{pmatrix} x \\ y\\z \end{pmatrix}\\\iff\begin{cases}x-3y+3z=4x\\3x-5y+3z=4y\\6x-6y+4z=4z\end{cases}\\\iff\begin{cases}-3x-3y+3z=0\\3x-9y+3z=0\\6x-6y=0\end{cases}\\\iff\begin{cases}x=y\\z=2x\end{cases}$$
So the eigenspace associated to $4$ is $\text{Vect}\left(\begin{pmatrix} 1 \\ 1\\2 \end{pmatrix}\right)$.
You can do the same for the eigenspace associated to $-2$.
