I have four nonlinear equations I want to find the points of intersection of these equations, and I used the software Mathematica, unfortunately after many hours of waiting it does not give me any result
Do you have an idea how to solve this kind of problem?.


My equations are the following 


1) $(52 \alpha ^2+\alpha (104 \beta +440 \sqrt{3}-1071)+\beta (65 \beta
 +110 \sqrt{3}-752)-2 (52 \gamma +55 \sqrt{3}-376) \delta +(-52 \gamma -440 \sqrt{3}+1071) \gamma -65 \delta ^2)=0$


2) $52 \alpha ^2+\alpha (-52 \beta +80 \sqrt{3}-450)+2 \beta (13 \beta +10
 \sqrt{3}+38)-52 \gamma ^2+\gamma (52 \delta -80 \sqrt{3}+450)-2 \delta (13
 \delta +10 \sqrt{3}+38)=0$


3) $\beta ^2-\alpha ^3=0$


4) $\delta ^2-\gamma ^3=0$



Intersection Solutions of four nonlinear equations
This is an expansion of the comment by Noam D. Elkies. Indeed, we can rewrite your equations 3) and 4) as $(\alpha,\beta)=(x^2,x^3)$ and $(\gamma,\delta)=(y^2,y^3)$ for some $x,y$. Substituting into eqs. 1) and 2) these expressions of $\alpha,\beta,\gamma,\delta$ in terms of $x,y$, we reduce your system of four eqs. to that of the following two: $P\_1=0=P\_2$, where $P\_1,P\_2$ are certain polynomials in $x,y$. 


The resultant $\text{Res}\_y(P\_1,P\_2)$ vanishes iff, for a given value of $x$, the polynomials $P\_1,P\_2$ have a common root $y$. We find that $\text{Res}\_y(P\_1,P\_2)=0=\text{Res}\_x(P\_1,P\_2)$ identically for all $x$ and $y$. So, $P\_1$ and $P\_2$ need some preliminary cleaning. Indeed, we see that 
\begin{equation}
 Q\_1:=\frac{P\_1}{y-x},\quad Q\_2:=\frac{P\_2}{2(y-x)}
\end{equation}
are polynomials in $x,y$. Thus, 
\begin{equation}
 \text{$(x,x)$ is a solution to our system for any complex $x$.} \tag{\*}
\end{equation}


The resultants $R\_y:=\text{Res}\_y(Q\_1,Q\_2)$ and $R\_x:=\text{Res}\_x(Q\_1,Q\_2)$ are polynomials in $x$ and in $y$, respectively, each having $16$ distinct complex roots, say $x\_1,\dots,x\_{16}$ for $R\_y$ and $y\_1,\dots,y\_{16}$ for $R\_x$. Details of all calculations here, as well as the particular enumeration of $x\_1,\dots,x\_{16}$ and $y\_1,\dots,y\_{16}$, can be seen in the [Mathematica notebook](https://my.pcloud.com/publink/show?code=XZICnC7ZUdC9JGUcYuHwFWWjU1GDa5sxkW0k) or its [pdf image](https://my.pcloud.com/publink/show?code=XZACnC7ZIYL4DHRMyJmg2WOQYWe9vmmIL52y). 


So, for each $i\in[16]:=\{1,\dots,16\}$ there is at least one $j\in[16]$ such that the pair $(x\_i,y\_j)$ is a solution to the system $Q\_1=0=Q\_2$. Moreover, it is easy to see that for each $i\in[16]\setminus\{3\}$, there is at most one (and therefore the only one) $j=j\_i\in[16]$ such that the pair $(x\_i,y\_{j\_i})$ is a solution to the system $Q\_1=0=Q\_2$. The corresponding pairs $(i,j\_i)$ are $(1, 3), (2, 4), (4, 2), (5, 6), (6, 5), (7, 11), (8, 12), (9, 10), (10, 9), (11, 7)$, $(12, 8), (13, 14), (14, 13), (15, 16), (16, 15)$. For the exceptional value $i=3$, we have at most two values of $j\in[16]$ (namely, $j=1$ and $j=3$) such that the pairs $(x\_3,y\_j)$ may be solutions to the system $Q\_1=0=Q\_2$; in fact we see that these pairs $(x\_3,y\_j)$, equal $(0,\sqrt3)$ and $(0,0)$, are indeed solutions to the system $Q\_1=0=Q\_2$; the pair $(0,0)$ has already been accounted for by (\*). 


Thus, we have described all the $16$ pairs $(x,y)$ that are (in addition to the "trivial" pairs given by (\*)) solutions to the system $Q\_1=0=Q\_2$. All these solutions $(x,y)$ are now straightforward to transcribe into a complete set of solutions $(\alpha,\beta,\gamma,\delta)$ of the original system. 


