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Hello coders i was actually writing some codes to solve quadratic equation with JAVA, below is what i came up with.

public class QuadraticEquationSolver
{
public static void main(String[] args)
{
/*Suppose our Quadratic  Equation to be solved
* is 2x2 + 6x + 4 = 0 .
* (Assuming that both roots are real valued)
*
* General form of a Quadratic  Equation is
* ax2 + bx + c = 0 where 'a' is  not equal to 0
*
* Hence a = 2, b = 6 and c = 4.
*/
int a =2;
int b =6;
int c =4;
//Finding out the roots
double temp1 = Math.sqrt(b * b -4* a * c);
double root1 = (-b +  temp1) / (2*a) ;
double root2 = (-b -  temp1) / (2*a) ;
System.out.println( "The roots of the Quadratic Equation \"2x2 + 6x + 4 = 0\" are "+root1+" and "+root2);
}
}

D34lw4p:
Hello coders i was actually writing some codes to solve quadratic equation with JAVA, below is what i came up with.

public class QuadraticEquationSolver
{
public static void main(String[] args)
{
/*Suppose our Quadratic  Equation to be solved
* is 2x2 + 6x + 4 = 0 .
* (Assuming that both roots are real valued)
*
* General form of a Quadratic  Equation is
* ax2 + bx + c = 0 where 'a' is  not equal to 0
*
* Hence a = 2, b = 6 and c = 4.
*/
int a =2;
int b =6;
int c =4;
//Finding out the roots
double temp1 = Math.sqrt(b * b -4* a * c);
double root1 = (-b +  temp1) / (2*a) ;
double root2 = (-b -  temp1) / (2*a) ;
System.out.println( "The roots of the Quadratic Equation \"2x2 + 6x + 4 = 0\" are "+root1+" and "+root2);
}
}

You need to place brackets (I mean parentheses) around the -4 *a, so like this (-4 * a). Otherwise the equation will come out incorrectly

You will be needing the general formula, I'm sure of that. But I can't even see it in your code.

mbatuku1:
You will be needing the general formula, I'm sure of that. But I can't even see it in your code.

He has it: unless if you are referring to the ax^2 + bx + c

he has the code as follows:

double temp1 = Math.sqrt(b * b -4* a * c);
double root1 = (-b + temp1) / (2*a) ;
double root2 = (-b - temp1) / (2*a) ;

Weldone OP.

One thing I'll like to add is, you should be aware that b * b -4* a * c will not always return a positive number, such as imaginary roots. So at the barest minimum you need an if else in your code.

You mind coding v1.1 of your solver?

dotman2460:
Weldone OP.

One thing I'll like to add is, you should be aware that b * b -4* a * c will not always return a positive number, such as imaginary roots. So at the barest minimum you need an if else in your code.

You mind coding v1.1 of your solver?

He doesn't have his code setup to prompt the user for input. He already has pre-set values for his code. This looks like an introductory assignment to a class in my opinion.

That's right.

Fulaman198:
He doesn't have his code setup to prompt the user for input. He already has pre-set values for his code. This looks like an introductory assignment to a class in my opinion.

I can't see your code, but you have to pass arguements in your parenthesis in a defined form.

You've declared the math standard library and your OOP structure. You will have to calculate your discriminant properly with correct parenthesis:
X = (-b +or- SQRT((b^2) - (4 * a * c))) / 2 * a

Note: First calculate your discriminant.

A quadratic equation has what we call a determinant and depending on if its positive negative or zero.the equation can either have one real root two distinct roots or no roots at all can't remember exactly but I coded one in c# as an exercise.

This program isn't ok