how to find solution of equations based on bisection method

I'm trying to implement the bisection method for finding the solutions of an equation.

The form of the equation is like this:
pe^(-x) + qsin(x) + rcos(x) + stan(x) + tx^2 + u = 0 , with 0 ≤ p,r ≤ 20, -20 ≤ q,s,t ≤ 0 e -20 ≤ u ≤ 20

input example:

3
1. 0 0 0 0 -2 1
1 0 0 0 -1 2
1 -1 1 -1 -1 1

should give :

0.7071
Impossible
0.7554

I tried to implement this but I can't display the result with 4 decimal places, and I realized that a b and c are in the form of x.x with only one decimal place. I think the problem starts here. Any help would be very appreciated . Here's my code

class p07{
public static void main(String [] args){
    Scanner in = new Scanner(System.in);
    int n= in.nextInt();

    for (int i=0 ; i < n; i++)
        bss(in.nextInt(), in.nextInt(),in.nextInt(), in.nextInt(),in.nextInt(), in.nextInt());

}

public static void bss(int p, int q, int r, int s, int t, int u){
    double fa=0, fb=0, fc=0;

    boolean flag=true;

    double a=-20;
    double b=a;

    while(flag){

        fa=p*Math.exp(a) + q*Math.sin(a) + r*Math.cos(a) + s*Math.tan(a) + t*Math.pow(a,2) + u;
        fb=p*Math.exp(b+1) + q*Math.sin(b+1) + r*Math.cos(b+1) + s*Math.tan(b+1) + t*Math.pow(b+1,2) + u;

        a++;b++;

        if( (fa < 0 && fb > 0) || (fa > 0 && fb < 0) )
            flag=false;
    }

    System.out.println("a= "+a+", b= "+b);
    System.out.println("f(a)= "+fa+", f(b)= "+fb);

    int k=4;
    double c=0.000;

    while(k!=0){

        c = (a+b)/2;
        fa = p*Math.exp(a) + q*Math.sin(a) + r*Math.cos(a) + s*Math.tan(a) + t*Math.pow(a,2) + u;
        fc = p*Math.exp(c) + q*Math.sin(c) + r*Math.cos(c) + s*Math.tan(c) + t*Math.pow(c,2) + u;

        if( fa < fc)
            b=c;
        else
            a=c;
        k--;
        System.out.println("a= "+a+",b= "+b+", c= "+c);
    }
    double sol =p*Math.exp(c) + q*Math.sin(c) + r*Math.cos(c) + s*Math.tan(c) + t*Math.pow(c,2) + u;
    System.out.println(sol);
}

}

Solution

some errors:

the third one doesnt converge.

bad test: you should test if fa and fc have different signs=> if (fa*fc<0)
you should also test if you get a root
I simplified your first test with a break
sol is not a solution, but f(pseudo-root)

I dont understand your problem about decimals: a,b,c are double.

I replace the loop with 10 iterations.

so, it gives:

double fa=0, fb=0, fc=0;

double a=-20;
double b=a+1;

while(true)
    {
    fa=p*Math.exp(a) + q*Math.sin(a) + r*Math.cos(a) + s*Math.tan(a) + t*Math.pow(a,2) + u;
    fb=p*Math.exp(b) + q*Math.sin(b) + r*Math.cos(b) + s*Math.tan(b) + t*Math.pow(b,2) + u;

    if( (fa < 0 && fb > 0) || (fa > 0 && fb < 0) )
        break;

    a++;b++;
    }

System.out.println("a= "+a+", b= "+b);
System.out.println("f(a)= "+fa+", f(b)= "+fb);

int k=10;
double c=0.0;

while(k!=0)
    {

    c = (a+b)/2;
    fa = p*Math.exp(a) + q*Math.sin(a) + r*Math.cos(a) + s*Math.tan(a) + t*Math.pow(a,2) + u;
    fc = p*Math.exp(c) + q*Math.sin(c) + r*Math.cos(c) + s*Math.tan(c) + t*Math.pow(c,2) + u;

    // ROOT
    if (fc==0) break;

    if (fa*fc<0)
        b=c;
    else
        a=c;

    k--;
    System.out.println("a= "+a+",b= "+b+", c= "+c+" f(c)"+fc);
}
double sol =p*Math.exp(c) + q*Math.sin(c) + r*Math.cos(c) + s*Math.tan(c) + t*Math.pow(c,2) + u;
System.out.println("root="+c);