Quadratic functions in c - quadratic

Question

I have almost everything working except for solving for X in line 25 i keep getting an error saying " term does not evaluate to a function taking 1787 arguments" i had it giving me a 1 or a 0 but as i kept messing with it i lost where i was at and saved over the copy. still new to posting sorry if its hard to read
#include <stdio.h>
#include <math.h>
void quadratic_function()
{
int a,b,c; // variables
long int result; // my X in the quadractic function
long int y,x; // the result
long int quadratic;
printf("enter values for a,b,c\n");
scanf("%i\n %i\n %i", &a,&b,&c);
printf("A=%i B=%i C=%i\n", a,b,c); //Displays Variables
y= pow(b, 2);
result= (y)*-4*(a)*(c); // b^2-4ac
printf("\n%li\n",result);
if (result<0)
printf("Imaginary Number"); // if negative
else (result>0);
x=(-b/2*(a)) +- (sqrt(pow(b, 2)) (-4*(a)*(c))) / (2*(a));
//solving for x
printf("\n %li\n",x);
a = a*x;
b = b*x;
quadratic=pow(a, 2)*(b)*(c); // if positive
//printf("Quadratic equation equal to %li",quadratic); // result
}
int main()
{
quadratic_function();
return 0;
}

Answer 1

The first thing I noticed is that you were trying to do the + and - portions of the quadratic equation at the same time. The equation
x = (-b +- sqrt(b^2 - 4ac)) / 2a
means the same as
x = (-b + sqrt(b^2 - 4ac)) / 2a AND x = (-b - sqrt(b^2 - 4ac)) / 2a
In other words, the equation has two answers if b^2 - 4ac is greater than 0, one answer if it is 0, and no answer if it is negative.
Another thing, the line else (result>0); doesn't really do anything. The rest of the code after that will execute even if you get b^2 - 4ac < 0
Finally, I wasn't entirely sure about your groupings or C++'s precedence with the negative sign, so I changed your parentheses around a bit.
y = pow(b, 2);
result = (y) - (4*a*c); // b^2-4ac
printf("\n%li\n", result);
if (result < 0) {
printf("Imaginary Number"); // if negative
} else if (result == 0) {
x = (-b) / (2 * a); // sqrt(0) = 0, so don't bother calculating it
a = a*x;
b = b*x;
quadratic=pow(a, 2)*(b)*(c);
printf("Quadratic equation equal to %li",quadratic); // result
} else if (result > 0) {
// solve for (-b + sqrt(b^2 - 4ac)) / 2a
x = ((-b) + sqrt(pow(b, 2) - (4 * a * c))) / (2 * a);
printf("\n %li\n",x);
a = a*x;
b = b*x;
quadratic=pow(a, 2)*(b)*(c);
printf("Quadratic equation equal to %li",quadratic); // result
// do it again for (-b - sqrt(b^2 - 4ac)) / 2a
x = ((-b) - sqrt(pow(b, 2) - (4 * a * c))) / (2 * a);
printf("\n %li\n",x);
a = a*x;
b = b*x;
quadratic=pow(a, 2)*(b)*(c);
printf("Quadratic equation equal to %li",quadratic);
}