A calculator will report $\lambda=0$ as an eigenvalue of algebraic multiplicity of 2, and $\lambda=-1$ as an eigenvalue of algebraic multiplicity 2 as well. Since eigenvalues are roots of the characteristic polynomial (Theorem EMRCP) we have the factored version
\begin{equation*}
\charpoly{A}{x}=(x-0)^2(x-(-1))^2=x^2(x^2+2x+1)=x^4+2x^3+x^2
\end{equation*}
The eigenspaces are then
\begin{align*}
\lambda&=0\\
A-(0)I_4&=
\begin{bmatrix}
19 & 25 & 30 & 5 \\
-23 & -30 & -35 & -5 \\
7 & 9 & 10 & 1 \\
-3 & -4 & -5 & -1
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & -5 & -5 \\
0 & \leading{1} & 5 & 4 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}\\
\eigenspace{A}{0}&=\nsp{C-(0)I_4}=
\spn{\set{\colvector{5\\-5\\1\\0},\,\colvector{5\\-4\\0\\1}}}
\end{align*}
\begin{align*}
\lambda&=-1\\
A-(-1)I_4&=
\begin{bmatrix}
20 & 25 & 30 & 5 \\
-23 & -29 & -35 & -5 \\
7 & 9 & 11 & 1 \\
-3 & -4 & -5 & 0
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & -1 & 4 \\
0 & \leading{1} & 2 & -3 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}\\
\eigenspace{A}{-1}&=\nsp{C-(-1)I_4}=
\spn{\set{\colvector{1\\-2\\1\\0},\,\colvector{-4\\3\\0\\1}}}
\end{align*}
Each eigenspace above is described by a spanning set obtained through an application of Theorem BNS and so is a basis for the eigenspace. In each case the dimension, and therefore the geometric multiplicity, is 2.
For each of the two eigenvalues, the algebraic and geometric multiplicities are equal. Theorem DMFE says that in this situation the matrix is diagonalizable. We know from Theorem DC that when we diagonalize $A$ the diagonal matrix will have the eigenvalues of $A$ on the diagonal (in some order). So we can claim that
\begin{equation*}
D=
\begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}
\end{equation*}