find root of a equation using newton raphson method

/*This program finds the root of the equations (x^5)+(x^4)+(x^3)+(x^2)+(x^1)+5=0
and 5(x^2)+2exp(-2x)=0 numerically using the Newton-Raphson method*/

#include <stdio.h>
#include <conio.h>
#include <math.h>


float NRiter(float);
float func(float);
float diff(float);
int c=1, count=0;

int main()
{
    float x, r;
   
   
    printf("For the equation (x^5)+(x^4)+(x^3)+(x^2)+(x^1)+5=0 \n\n");
    start:
    printf("x0 = ");
    scanf("%f", &x);
    r = NRiter(x);
    if(r==0) printf("This equation does not have a real solution");
    else printf("The root is %f\n\n\n", r);
   
    if(c==1)
      {
         printf("For the equation 5(x^2)+2exp(-2x)=0 \n\n");
         c=2;
         goto start;
      }
   
    getch();
    return 0;
}
   
float NRiter(float x)
{
    float t, r, d;
    ++count;
    r = x - func(x)/diff(x);
    t = fabs(func(r));
    if(count>100) return 0;   
    if(t<0.001) return r;
    else
       {
         d = NRiter(r);
         return d;
       }
}
      
float func(float x)
{
    float y;
    if(c==1) y = pow(x,5)+pow(x,4)+pow(x,3)+pow(x,2)+pow(x,1)+5;
    else y = 5*pow(x,2)+2*exp(-2*x);
    return y;
}      
   
float diff(float x)
{
    float d;
    if(c==1) d = 5*pow(x,4)+4*pow(x,3)+3*pow(x,2)+2*pow(x,1)+1*pow(x,0);    
    else d = 10*x-4*exp(-2*x);
    return d;
}