how to put in mathematical equation in C - c

Question

I've been trying to look up on Google how to put in an equation in my program but wasn't able to find any. How do you include:
x = ( -b + √b2 - 4ac ) / 2a
in the program?
Here's my code:
{
int a, b, c;
float x;
//statements
printf("Enter three integers: ");
scanf("%d %d %d", &a, &b, &c);
//computeforX
x = ( -b + √b2 - 4ac ) / 2a
printf("The value of x is %.1f", x);
return 0;
}

Answer 1

Assuming we're talking about C (or C++) here, you will need to investigate the sqrt function, and maybe also the pow function as well (although that's unnecessary because b-squared can be computed as b*b).
Note that you will need to convert all of your input values to float or double before you start the calculation, otherwise you will not get the intended result.

Answer 2

You need a table to allow you to translate:
a+b -> a+b
a-b -> a-b
a/b -> a/b
ab -> a*b
√x -> sqrt(x)
x² -> x*x (If you want to square something more complicated it might be best to use a temporary variable for the value to be squared, breaking your equation up into pieces.)
Note that if you divide an int by an int in C you get an int. So better convert those ints to doubles before dividing.

Answer 3

If we are dealing with C++ it would be something like
#include <iostream.h>
#include <cmath>
int main ()
{
//Declare Variables
double x,x1,x2,a,b,c;
cout << "Input values of a, b, and c." ;
cin >>a >>b >>c;
if ((b * b - 4 * a * c) > 0)
cout << "x1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a)" &&
cout << "x2 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a)";
if else ((b * b - 4 * a * c) = 0)
cout << "x = ((-b + sqrt(b * b - 4 * a * c)) / (2 * a)"
if else ((b * b - 4 * a * c) < 0)
cout << "x1 = ((-b + sqrt(b * b - 4 * a * c) * sqrt (-1)) / (2 * a) &&
cout << "x2 = ((-b + sqrt(b * b - 4 * a * c) * sqrt (-1)) / (2 * a);
return (0);
}
Now why do i have this wierd feeling I just did someone's first semester programming class' homework?
Granted its been years and I don't even know if that will compile but you should get the idea.

Answer 4

I am really depressed looking the quality of above answers and help, which has been given.
I hope to improve the content of this thread.
One can compile the C file below with the command line gcc file.c -o file -lm.
Herewith a possible solution in C:
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
int main(){
int da, db, dc;
double x, a,b,c;
//statements
printf("Enter three integers: ");
scanf("%d %d %d", &da, &db, &dc);
a = (double)da;
b = (double)db;
c = (double)dc;
//computeforX
x = (double) ( -b + sqrt(b * b) - 4 * a * c ) / ( 2 * a ) ;
printf("The value of x is %g \n", x);
return 0;
}