URI 1036 solution

#include<stdio.h>

#include<math.h>

 int main()

{

  double A,B,C,R1,R2,D;

  scanf("%lf%lf%lf",&A, &B, &C);

  D = ((B*B) - (4*A*C));

  if(A!=0 && D>0)

  {

      R1 = (- B + sqrt(D))/(2*A);

      R2 = (- B - sqrt(D))/(2*A);

      printf("R1 = %.5lf
R2 = %.5lf
",R1,R2);

  }

  else{

      printf("Impossivel calcular
");

  }

 return 0;

}