Numerical Mathematics : Newton Rapson Solution (C Programming )

Sample equation for Newton Rapson

where X=3;

F(x)=2x³-11.7x²+17.7x¹-5

 //NETON RAPSON  
 #include<stdio.h>  
 #include<math.h>  
 #include<iostream>  
 using namespace std;  
 int main()  
 {  
   float xi,Es,a[40],xiii,old=0,Ea;  
   int n,K=0;  
   cout<< "HIghest power in the equation : " ;  
   cin>>n;  
   //.........................................  
   printf("Enter Xi =");  
   scanf("%f",&xi);  
   printf("Enter Es =");  
   scanf("%f",&Es);  
   //..........................................  
   printf("\n");  
   printf("enter coefficient :\n");  
   for (int i=n; i>=0; i--)  
   {  
     printf("X^%d =",i);  
     scanf("%f",&a[i]);  
   }  
   //...........................................  
   while(1)  
   {  
     float fxi=0,deff=0;  
     printf("\n\n\nSTEP no: %d\n",K+1);  
     for (int i=n; i>=0; i--)  
     {  
       fxi=fxi+a[i]*pow(xi,i);  
     }  
     for(int i=n; i>=0; i--)  
     {  
       deff=deff+i*a[i]*pow(xi,i-1);  
     }  
     printf("Xi  = %.2f",xi);  
     printf("\nfxi = %.2f",fxi);  
     printf("\ndeff  = %.2f",deff);  
     xiii=(xi-(fxi/deff));  
     printf("\nnew xi  = %.2f",xiii);  
     Ea=((xiii-old)/xiii)*100;  
     if(Ea<0)  
     {  
       Ea=Ea*(-1);  
     }  
     printf("\nEa= %f",Ea);  
       old=xiii;  
       if(Ea<Es)  
     {  
       break;  
     }  
     xi=xiii;  
     K++;  
   }  
   return 0;  
 }