A cubic over $k$ in Weierstrass form (affine form) is given by $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.$$ The discriminant is defined by $$\Delta = -b_2^2b_8-8b_4^3-27b_6^2+9b_2b_4b_6,$$ where $ b_2=a_1^2+4a_2$, $ b_4=2a_4+a_1a_3 $, $b_6=a_3^2+4a_6$ and $b_8 = a_{1}^{2} a_{6}+4 a_{2} a_{6}-a_{1} a_{3} a_{4}+a_{2} a_{3}^{2}-a_{4}^{2}$.
Finally, a elliptic curve over $k$ is a cubic in Weiertrass form, where $\Delta \neq 0$ (i.e., is a non-singular cubic in Weiertrass form).
We can make some substitutions to simplify the equation of cubic in Weiertrass form, the first assumes $char(k)$ is not $2$. Replacing $y$ by $ \frac{1}{2} \left(y-a_1x-a_3\right)$, the result is $$y^2=4x^3+b_2x^2+2b_4x+b_6.$$
The second assumes in addition that $char(k) \neq 3$. Replace $(x,y)$ by $\left( \frac{x-3b_2}{36}, \frac{y}{108} \right) $, and the result is $$y^2=x^3-27c_4-54c_6,$$ where $c_4=b_2^2-24b_ 4$ and $c_6=-b_2^3+36b_2b_4-216b_6.$
Moreover, when $char(k)$ not is $2$ or $3$, we have $$ 1728\Delta=c_4^3 - c_6^2. $$
Now, consider the cubic $y^2=x^3+ax+b$ over $k$. If $char(k)$ not is $2$ or $3$, we have $c_4=-48a $ and $c_6=-864b$, so $$\Delta = \frac{(-48a)^3-(-864)^2}{1728} = -16(4a^3+27b^2).$$
Thus, assuming that $char(k)\neq2$ and $char(k) \neq 3$, an elliptic curve over $k$ is given by $$y^2=x^3+ax+b,$$ where $\Delta=-16(4a^3+27b^2) \neq 0$.
Note that, $\Delta=-16(4a^3+27b^2) = 0$ if, and only if, $4a^3+27b^2 = 0$, because $16=2^4 \neq 0$ in $k$ with $char(k) \neq 2$. Thus, the factor $−16$ is irrelevant in this case.
See Chapter III of the book 'Elliptic Curves' of Anthony Knapp for more information.
