Need to draw phase portrait near the equilibrium points of differential equation So, this equation 
$$\ddot{x}+3\dot{x}-4x+2x^2 = 0.$$
I can write like a system
\begin{equation}
\left\{ \begin{array}{ll}
\dot{x} = v, \\
\dot{v} = 2x^2 - 4x - 3v.
\end{array} \right.
\end{equation}
Finding the equilibrium points:
\begin{equation}\nonumber
\left\{ \begin{array}{ll}
v = 0 \\
2x^2 - 4x - 3v = 0
\end{array} \right. \ \ \Rightarrow 2x^2 - 4x = 0,  \ \ \Rightarrow  \ \ x_1 = 0, \ x_2 = 2. \ \ \Rightarrow \\ \Rightarrow \boldsymbol{(0,0)}, \
\boldsymbol{(2,0)} - \text{equilibrium  points}
\end{equation}
The matrix for the system
\begin{equation} 
A(x,y)=\left(\begin{array}{ccc}
0     &   1 \\
4-2х  &  -3 \\
\end{array}
\right)\end{equation}
Now I need to determine what is happening in each equilibrium point.
For example, first point, $\boldsymbol{(0,0)}$:
$$A(0,0)=\left(\begin{array}{ccc}
0   &  1 \\
4   &  -3 \\
\end{array}
\right)$$
$$\lambda^2 + 3\lambda -4 = 0,$$
$$\lambda_1=1, \ \ \lambda_2=-4$$
Here everything looks fine, the different signs of real roots saying to me it's saddle.
But then, let's look at the point $\boldsymbol{(2,0)}$:
$$A(2,0)=\left(\begin{array}{ccc}
0   &  1 \\
0   &  -3 \\
\end{array}
\right)$$
$$\lambda^2 - 3\lambda = 0,$$
$$\lambda_1 = 0, \ \ \lambda_2 = -3.$$
It means just one eigen vector. Would it be a sink? On a picture you can see numerical solution:

I think, after I write it here, I get it why the picture should be like this. But I wrote so a lot, so I will post it, maybe someone can explain it better or add something. Thank you for attention.

Your first calculation is correct and leads to an unstable saddle.
However, you made a slight error in your Jacobian which affected your second calculation (the first was not affected due to the zero value).
For the Jacobian, you should have:
$$J(x,y)=\begin{bmatrix}
0     &   1 \\
4-4х  &  -3 \\
\end{bmatrix}$$
When you evaluate the Jacobian at your second critical point, $(2,0)$, you end up with the eigenvalues:
$$\lambda_{1,2} = \dfrac{1}{2}\left(-3 ~ \pm ~ i \sqrt{7}\right)$$
This tells you that you have a stable spiral at $(2,0)$.
Your phase portrait agrees with this. Here it is plotted with a different tool:

I would also like to mention that you are mixing variable names in your reduction and in other places in the calculations and that will surely confuse you and potentially lead to errors, so you should clean that up.
For example, you could have written the system as:
$$x' = y \\ y' = 4 x - 2 x^2 - 3y$$
Notice how that makes everything else jive.
