Finding a nullspace of a matrix - what should I do after finding equations? I am given the following matrix $A$ and I need to find a nullspace of this matrix.

$$A =
 \begin{pmatrix}
  2&4&12&-6&7 \\
  0&0&2&-3&-4 \\
  3&6&17&-10&7
 \end{pmatrix}$$

I have found a row reduced form of this matrix, which is:
$$A' =
 \begin{pmatrix}
  1&2&0&0&\frac{23}{10} \\
  0&0&1&0&\frac{13}{10} \\
  0&0&0&1&\frac{22}{10}  \end{pmatrix}$$
And then I used the formula $A'x=0$, which gave me:
$$A' =
 \begin{pmatrix}
  1&2&0&0&\frac{23}{10} \\
  0&0&1&0&\frac{13}{10} \\
  0&0&0&1&\frac{22}{10} 
 \end{pmatrix}
\begin{pmatrix}
  x_1 \\
  x_2 \\
  x_3  \\
  x_4 \\
  x_5
 \end{pmatrix}=
\begin{pmatrix}
  0 \\
  0 \\
  0  
 \end{pmatrix}$$
Hence I obtained the following system of linear equations:
$$\begin{cases} x_1+2x_2+\frac{23}{10}x_5=0 \\ x_3+\frac{13}{10}x_5=0 \\ x_4+\frac{22}{10}x_5=0 \end{cases}$$
How should I proceed from this point?
Thanks!

\begin{cases} x_1+2x_2+\frac{23}{10}x_5=0 \\ x_3+\frac{13}{10}x_5=0 \\ x_4+\frac{22}{10}x_5=0 \end{cases}
$$
x_1=-2x_2-\dfrac{23}{10}x_5
$$
$$
x_3=-\dfrac{13}{10}x_5
$$
$$
x_4=-\dfrac{11}{5}x_5
$$
Therefore,basis of null space=
$$
\begin{pmatrix}
  -2 \\
  1 \\
  0  \\
  0 \\
  0
 \end{pmatrix},
\begin{pmatrix}
  -\dfrac{23}{10} \\
  0 \\
  -\dfrac{13}{10}  \\
  -\dfrac{11}{5} \\
  1
 \end{pmatrix}
$$
