c program for solving quadratic equation | CTechnotips

1. C program to calculate roots of a quadratic equation

2. Quadratic equation in c language

#include<stdio.h>

#include<math.h>

int main(){

  float a,b,c;

  float d,root1,root2;  

 

  printf("Enter a, b and c of quadratic equation: ");

  scanf("%f%f%f",&a,&b,&c);

   

  d = b * b - 4 * a * c;

  

  if(d < 0){

    printf("Roots are complex number.\n");

    printf("Roots of quadratic equation are: ");

    printf("%.3f%+.3fi",-b/(2*a),sqrt(-d)/(2*a));

    printf(", %.3f%+.3fi",-b/(2*a),-sqrt(-d)/(2*a));

  

    return 0; 

  }

  else if(d==0){

   printf("Both roots are equal.\n");

   root1 = -b /(2* a);

   printf("Root of quadratic equation is: %.3f ",root1);

   return 0;

  }

  else{

   printf("Roots are real numbers.\n");

  

   root1 = ( -b + sqrt(d)) / (2* a);

   root2 = ( -b - sqrt(d)) / (2* a);

   printf("Roots of quadratic equation are: %.3f , %.3f",root1,root2);

  }

  return 0;

}

Sample output:

Enter a, b and c of quadratic equation: 2 4 1

Roots are real numbers.

Roots of quadratic equation are: -0.293, -1.707

1. How to find a b and c in a quadratic equation

#include<stdio.h>

#include<math.h>

int main(){

  float a,b,c;

  float d,root1,root2;  

  printf("Enter quadratic equation in the format ax^2+bx+c: ");

  scanf("%fx^2%fx%f",&a,&b,&c);

   

  d = b * b - 4 * a * c;

  

  if(d < 0){

    printf("Roots are complex number.\n");

   

    return 0;

  }

 

   root1 = ( -b + sqrt(d)) / (2* a);

   root2 = ( -b - sqrt(d)) / (2* a);

   printf("Roots of quadratic equation are: %.3f , %.3f",root1,root2);

  return 0;

}

Sample output:

Enter quadratic equation in the format ax^2+bx+c: 2x^2+4x+-1

Roots of quadratic equation are: 0.000, -2.000