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A simple program to solve quadratic equations with

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4.00/5 (7 votes)
8 Nov 2010CPOL 38K   3   7
Simple and prints imaginary roots too!float a,b,c,x1,x2,d,dsq;printf("ax^2 + bx + c = 0");printf("\nEnter a,b,c separated by commas : \n");scanf("%f,%f,%f",&a,&b,&c);d = b*b-(4*a*c);if(d>=0){dsq=sqrt(d);x1 = (-b+dsq)/(2*a);x2 = (-b-(dsq))/(2*a);printf("\nRoot 1 : %f\nRoot 2...
Simple and prints imaginary roots too!
C#
float a,b,c,x1,x2,d,dsq;
printf("ax^2 + bx + c = 0");
printf("\nEnter a,b,c separated by commas : \n");
scanf("%f,%f,%f",&a,&b,&c);
d = b*b-(4*a*c);
if(d>=0)
{
dsq=sqrt(d);
x1 = (-b+dsq)/(2*a);
x2 = (-b-(dsq))/(2*a);
printf("\nRoot 1 : %f\nRoot 2 : %f",x1,x2);
}
if(d<0)
{
d = ((4*a*c)-pow(b,2))/(2*a);
printf("\nRoot 1 : %f+%fi",((-b)/(2*a)),d);
printf("\nRoot 2 : %f-%fi",((-b)/(2*a)),d);}

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)


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Comments and Discussions

 
SuggestionYou have missed one Pin
VISWESWARAN199814-Jun-16 20:18
professionalVISWESWARAN199814-Jun-16 20:18 
GeneralReason for my vote of 2 Can crash too ! Pin
YvesDaoust8-Nov-10 23:27
YvesDaoust8-Nov-10 23:27 
Generalthats the most basic rule in the book Pin
Sherylee8-Nov-10 22:38
Sherylee8-Nov-10 22:38 
General1) It doesn't matter what the real world quadratic equations... Pin
Ron Beyer8-Nov-10 4:34
professionalRon Beyer8-Nov-10 4:34 
GeneralThank you for the suggestion.. I do not wish to be rude. If... Pin
Anshul R8-Nov-10 0:13
Anshul R8-Nov-10 0:13 
GeneralReason for my vote of 2 Inefficient and does not handle divi... Pin
Andrew Phillips7-Nov-10 15:19
Andrew Phillips7-Nov-10 15:19 
GeneralThis is better than the others as you avoid the domain error... Pin
Andrew Phillips7-Nov-10 15:14
Andrew Phillips7-Nov-10 15:14 

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