Code to find quadratic roots fail when huge numbers are used

I got this problem from an online course and here I had to write a small program to find quadratic roots, and the return type should be Set<Integer>. I am still learning Java and still not familiar working with those types.

I think everything is not wrong until this part,

if(discriminant > 0) {
        root1 = (int)(-b + Math.sqrt(discriminant)) / (2 * a);
        root2 = (int)(-b - Math.sqrt(discriminant)) / (2 * a);
        result.add(root1);
        result.add(root2);
    }

As I have to return the final roots as a Set<Integer> type I had to force convert double to int returned by 'Math.sqrt'. I am not sure if this is what causing the issues. And if so I am not sure if how to solve this, because I can't add double values to a set<Integer>.

I tested this code with few test cases, and it failed when using really big values, like ~2,000,000,000 for c.

And this is the code I came up with so far.

public class Quadratic {


public static Set<Integer> roots(int a, int b, int c) {
    int root1;
    int root2;
    int discriminant = b * b - 4 * a * c;
    Set<Integer> result = new HashSet<Integer>(); 
    
    if(discriminant < 0) {
      String rootsAreImaginary = "Roots are imaginary";
      System.out.println(rootsAreImaginary);
    }
    if(discriminant == 0) {
        root1 = (-b) / (2 * a);
        root2 = root1;
        result.add(root1);
        result.add(root2);
    }
    if(discriminant > 0) {
        root1 = (int)(-b + Math.sqrt(discriminant)) / (2 * a);
        root2 = (int)(-b - Math.sqrt(discriminant)) / (2 * a);
        result.add(root1);
        result.add(root2);
    }
    return result;
}

If there are better ways to do this, please feel free to show me. Thank you so much in advance.

You can use BigDecimal or BigInteger to do the computations. These would have to be stored in a Set of the proper type, e.g. Set<BigDecimal>.

Both of those classes have methods to return the value of the related primitive (BigDecimal#doubleValue() and BigInteger#longValue()). But precision and size concerns still apply as you may not be able to fit the result into the class's related primitive.