// C program to demonstrate
// double and float precision values

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

// utility function which calculate roots of
// quadratic equation using double values
void double_solve(double a, double b, double c){
    double d = b*b - 4.0*a*c;
    double sd = sqrt(d);
    double r1 = (-b + sd) / (2.0*a);
    double r2 = (-b - sd) / (2.0*a);
    printf("%.5ft%.5fn", r1, r2);
}

// utility function which calculate roots of
// quadratic equation using float values
void float_solve(float a, float b, float c){
    float d = b*b - 4.0f*a*c;
    float sd = sqrtf(d);
    float r1 = (-b + sd) / (2.0f*a);
    float r2 = (-b - sd) / (2.0f*a);
    printf("%.5ft%.5fn", r1, r2);
}

// driver program
int main(){
    float fa = 1.0f;
    float fb = -4.0000000f;
    float fc = 3.9999999f;
    double da = 1.0;
    double db = -4.0000000;
    double dc = 3.9999999;

    printf("roots of equation x2 - 4.0000000 x + 3.9999999 = 0 are : n");
    printf("for float values: n");
    float_solve(fa, fb, fc);

    printf("for double values: n");
    double_solve(da, db, dc);
    return 0;
}

roots of equation x2 - 4.0000000 x + 3.9999999 = 0 are : 
for float values: 
2.00000    2.00000
for double values: 
2.00032    1.99968