quadratic equation solving

i use codeblocks for compiling.
The following is the code that i have written for solving quadratic equation.

Code:

---


#include<math.h>


main()
{

    float a,b,c,discriminant,root1,root2;
    printf("input values of a,b, and c\n");
    scanf("%f %f %f",&a,&b,&c);
    discriminant = (b*b - 4*a*c);
    if (discriminant<0)
      printf("\n\nROOTS ARE IMAGINARY\n");
    else
    {
        root1=(-b+sqrt(discriminant))*1/(2.0*a);
        root2=(-b-sqrt(discriminant))*1/(2.0*a);
        printf("\n\nRoot1=%f\n\nRoot2=%f\n,root1,root2");

    }

}

---

but when i enter the numbers 2,4,-16
it shows
root1=0.000000
root2=0.000000

i dont know what is wrong can anybody please help

Code:

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printf("\n\nRoot1=%f\n\nRoot2=%f\n,root1,root2");

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Do you see anything wrong with this statement? Pay attention to the warnings your compiler emits. Also, include <stdio.h>

Include stdio.h which include prototypes of printf,scanf,...
main function signature should be int main(void) and return 0 at the end.

Code:

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int main(void)
{
  ....
 return 0;    // success
}

---

Code:

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        printf("\n\nRoot1=%f\n\nRoot2=%f\n,root1,root2");
        printf( format_string,  ....);
  eg. printf("int %d\n", 3);
        printf("root = %f\n",root);

---

what is the error with printf statement.can you please tell

thank you @Bayint Naung.It was foolish of me to write that line incorrectly.I dont know why the compiler did not point out.

It's pointing you out 'saying not enough arguments..... because %f %f in your format string but no args.'

yes i got it now