0

I been working on this problem for so long, when I compile, it give me an error message that I misplace else at "else if (a==0 && b==0 && c!=0)"

when I remove the else at that line, it mess up my code, it won't display like it suppose to , but it display "floating point error" , that mean it doesn't divide by zero, but I need to override dividing by zero in order to display the messages that I want it to display in my code.

please help.

```
#include <stdio.h>
#include <math.h>
int main()
{
float a;
float b;
float c;
float d;
float e;
float f;
float x1;
float x2;
int opselect;
float sqr;
char more = 'y';
while (more == 'y' || more == 'Y')
{
clrscr();
printf ("\n\n\t\t\tSolving Quadratic Equation");
printf ("\n\n\t\t\tPlease enter a number for A: ");
scanf ("%f", &a);
printf ("\n\n\t\t\tPlease enter a numner for B: ");
scanf ("%f", &b);
printf ("\n\n\t\t\tPlease enter a number for C: ");
scanf ("%f", &c);
if (a==0 && b==0 && c==0) {
opselect = 1;
else if (a==0 && b==0 && c!=0)
opselect = 2;
else if (a==0 && b!=0 && c!=0)
opselect = 3;
else if (sqr < 0)
opselect = 4;
else if (sqr == 0)
opselect = 5;
else
opselect = 6;
}
switch (opselect)
{
case 1:
printf ("\n\n\t\t\tInfinite Solution!!");
break;
case 2:
printf ("\n\n\t\t\tContradict Solution!!");
break;
case 3:
x1 = -b/a;
printf ("\n\n\t\t\tSingle Root\n\t\t\tx= %4.2f", x1);
break;
case 4:
d = (sqrt(b*b) - (4.0*a*c));
e = -b/(2*a);
f = d/(2*a);
printf ("\n\n\t\t\t\tTwo complex root\n\n\t\t\t x1 = %4.2f + %4.2fi", e, f);
printf ("\n\n\t\t\t x2 = %4.2f - %4.2fi", e, f);
break;
case 5:
x1 = (-b + (sqrt(b*b) - (4.0*a*c))) / (2.0 * a);
printf ("\n\n\t\t\tRepreated Root\n\t\t\tx = %4.2f", x1);
break;
case 6:
x1 = (-b + (sqrt(b*b) - (4.0*a*c))) / (2.0 * a);
x2 = (-b - (sqrt(b*b) - (4.0*a*c))) / (2.0 * a);
printf ("\n\n\t\t\t\tTwo real root\n\n\t\t\t x1 = %4.2f\t\tx2 = %4.2f", x1, x2);
break;
}
printf ("\n\n\t\t\tDo more (Y/N) ? ");
scanf ("%s", &more);
}
return 0;
}
```