java: quadratic equation coefficients

Question

Can you guys explain why the code don't work? :(
Help please

ax2 + bx + c = 0

-> return one or two roots of the equation if there is any in the set of real numbers
-> Input value via System.in
-> Output value must be printed to System.out

Output format is:
-> "x1 x2" (two roots in any order separated by space) if there are two roots
-> "x" (just the value of the root) if there is the only root
-> "no roots" (just a string value "no roots") if there is no root
-> The a parameter is guaranteed to be not zero

import java.util.Scanner;

//import static java.lang.Math.sqrt;

public class QuadraticEquation {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
double a = scanner.nextDouble();
double b = scanner.nextDouble();
double c = scanner.nextDouble();

double discriminant = Math.pow(b, 2) - 4 * a * c;

System.out.print("The equation has ");
if (discriminant > 0)
{
double root1 = (-b + Math.pow(discriminant, 0.5)) / (2 * a);
double root2 = (-b - Math.pow(discriminant, 0.5)) / (2 * a);
System.out.println("two roots " + root1 + " and " + root2);
}
else if (discriminant == 0)
{
double root1 = (-b + Math.pow(discriminant, 0.5)) / (2 * a);
System.out.println("one root " + root1);
}
else
System.out.println("no real roots");
}
}

I need to receive (for example):

Input: 2, 5, -3
Output: -3 0.5

Input: 2, 2, 2
Output: no roots

Answer 1

Simple, Scanner use a delimiter which is a regex that acts as a separator, the problem is that the delimiter is by default settled to "\s+" (spaces), so the comma is considered as an invalid character(like a, b, ... , y, z...), and to fix this, just replace the delimiter by another regex with this :
Scanner scanner = new Scanner(System.in).useDelimiter("[,\s\n]\s*");

Answer 2

add this after create Scanner:
scanner.useDelimiter(
"(>?\s*,\s*)|\s+");

this solution is similar to VCoder, but maches also space before coma

Answer 3

if input is
2 5 -3
(without comas)

change + and - in ( -b - ... )
double root1 = (-b - Math.pow(discriminant, 0.5)) / (2 * a);
double root2 = (-b + Math.pow(discriminant, 0.5)) / (2 * a);

Answer 4

zeniak, the problem is that running the code will cause an error, even if change the + and - 😅

Answer 5

it works for me how can I simulate it ?

Answer 6

The error will occur if you input 2, 4, 6 (without my answer)

Answer 7

comas are necessary?

Answer 8

Yeah

Answer 9

edit