SRM 635 Div 2 250 & 500 Points

After long time not participate in topcoder because of work load, i decide after office today afternoon to participate in simple problem div 2

*IdentifyingWood (250 points) : *

for this case in my mind i need to 1 loop iterate through n ( length of s text ) and check if i can flag every char in t

public class IdentifyingWood {

    public static String check(String s, String t){
        int ls = s.length();
        int lt = t.length();
        int a = 0;
        String wood = "Yep, it's wood.";
        String none = "Nope.";
        for (int i = 0; i < ls; i++) {
            if (a < lt){
                if (s.charAt(i) == t.charAt(a)){
                    a++;
                }
            }
            else{
                break;
            }
        }
        if (a == lt){
            return wood;
        }
        return none;
    }

}

*QuadraticLaw (500 points) : *

in this case the solutison is simple by using quadratic equation :
** ( -b +- sqrt(b² - 4ac)/-2a**

But when i apply this formula to the solution Java Precition point is killing me. When the solution 999999998999999999l is 999999998 not 999999999 arghhhhhhhhhhhhhhh this is my first answer

public class QuadraticLaw {

    public static void main(String[] args) {
        System.out.println(getTime(999999998999999999l));
    }

    public static long getTime(long d){
        long bac = (1    + ( 4 * d) ); 
        long res1 = (long) (1 - Math.sqrt(bac))/-2;
        long res2 = (long) (1 + Math.sqrt(bac))/-2;

        return Math.max(res1, res2);
    }

}

But once again i found a bruteforce solution is by using a loop from a square of d and add it once by once :

public class QuadraticLaw {

    public static void main(String[] args) {
        System.out.println(getTime(999999998999999999l));
    }

    public static long getTime(long c){
        long t = (long) Math.sqrt(c);
        long res = t * (t+1);

        while(res <= c){
            t++;
            res = t * (t+1);
            if (res < c)return --t;
        }
        return t-1;
    }

}