how to solve discrete logarithm problem information requested I have seen your example about ecdlp solver :The Problem is as follows:

$$E\backslash GF(p):y^2=x^3+17230x+22699$$

where $p=23981$, point $G$ with prime order $|G| = 109$

Alice creates a public key by selecting a private key $d&lt;q$, public key $Q=[d]G = (3141,12767)$

Therefore;


public information : $a,b,p,G,q,Q$ 
private key : $d$


However, this curve has the following characteristic:

$\Delta= −16( 4a^3+27b^2) \bmod p =0$
That is, the discriminant is 0. and embedding degree is 2

I, however, don't understand where comes out the value  (23796,0) i.e how can I calculate it? and then the following formula equation: after which we obtain the corresponding curve
$$y^2=x^3+23426x^2$$


is it possible for a further explanation of how you determine the point $(27396,0)$?
Let $f(x,y) = -y^2 + x^3+17230x+22699$ over $\Bbb F_p$ with $p=23981$. A point on the curve is a singular point if and only if the partial derivatives are vanishes at that point. The partial derivatives are;


$\frac{\partial f}{\partial x} = 3x^2 + 17230 =0 \pmod p$ and vanishes at $x=\{185,23796\}$ found by WolframAlpha or it can be found by Tonelli-Shanks.
$\frac{\partial f}{\partial y} = -2y = 0 \pmod p$ and vanishes at $y=0$ 


The vanishing point $(185,0)$ is not on the curve, however, $(23796,0)$ is.

Therefore $(23796,0)$ is a singular point for the Curve.

Then we then translate the origin to this singular point $(23796,0)$, that is replace $(x,y)$ by $(x+23796,y+0$) in the equation of $E$, yielding the equation of the curve in the shifted referential:
$$y^2 = x^3 + 23426x^2$$
The rest is in the answer which the OP had trouble with.

Note: credit goes to @kelalaka for the better of the present answer, in particular introducing $f(x,y)$ and its partial derivatives.