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!

here is an almost algorithm to find a basis for the null space of a matrix $A:$
(a) row reduce $A,$ 
(b) identify the free and pivot b=variables. the variables corresponding to the non pivot columns, here they are $x_2$ and $x_4,$ are called the free variables and the rest are called pivot variables.
(c) you can set one of the free variables to one and the rest to zero. solve for the pivot variables. that gives you one solution in the null space; cycle through the free variables to find a basis for the null space.
i will show you how to  find one solution; you can find the other.
$x_2 = 1, x_4 = 0$  you can solve for $x_5$ first, giving you $x_5 = 0$ and then $x_3 = 0, x_1 = -2.$  so one vector in the basis is 
$$ \pmatrix{-2, &1, &0, &0, &0}^\top.$$
