Solve Quadratic Equation Using Numerical Approach

Quadratic.c

/*

Author: Redeem Grimm
Date: April 20
Purpose: Solves Quadratic Equation Using Numerical Approach

*/

//variable declaration
float a,b,c,rRoot,root1,root2,discriminant,compute1,compute2;

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int main(){

//prompting the user and getting input
printf("Enter The Three Constants: ");
scanf("%f%f%f",&a,&b,&c);
  
//discriminant
discriminant=b*b-4*a*c;

//Conditions Before Various Operations Are Carried Out
if(discriminant>0){
  
    //execute equalDistinctRoots
    equalDistinctRoots();
}else if(discriminant==0){
  
   //execute equalRoots
   equalRoots();
}else{
  
   //execute complexRoots
   complexRoots();
}
return 0;
}

//Methods To Carry Out Different Operations
void equalDistinctRoots(){
    root1=(-b + sqrt(discriminant))/(2*a);
    root2=(-b - sqrt(discriminant))/(2*a);
    printf("Status: Roots Are Real And Distinct.\n");
    printf("Root1: %.2f\n",root1);
    printf("Root2: %.2f\n",root2);
    }
void equalRoots(){
    rRoot=-b/(2*a);
    printf("Status: Roots Are Real And Equal.\n");
    printf("Root1 = Root2 = %.2f\n",rRoot);
    }
void complexRoots(){
    compute1=sqrt(-discriminant)/(2*a);
    rRoot=-b/(2*a);
    printf("Status: Roots Are Complex.\n");
    printf("Root1: %.2f + %.2fi\n",rRoot,compute1);
    printf("Root2: %.2f - %.2fi\n",rRoot,compute1);
}

So Simplified!