For this Question: Write a program to find the roots of a quadratic equation.
<b>Note</b>: Try to do it using Switch Case.

<b>Quadratic equations</b> are the polynomial equations of degree 2 in one variable of type f(x) = ax<sup>2</sup> + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where 'a' is called the leading coefficient and 'c' is called the absolute term of f (x).The first line of the input contains the three integer values a, b, and c of equation ax^2 + bx + c.

<b>Constraints</b>
1 &le; a, b, c &le; 50Print the two roots in two different lines and for imaginary roots print real and imaginary part of one root with (+/- and i )sign in between in one line and other in next line. For clarity see sample Output 2.

<b>Note</b> Imaginary roots can also be there and roots are considered upto 2 decimal places.Sample Input 1 :
4 -2 -10

Sample Output 2 : 
1.85
-1.35

Sample Input 2 : 
2 1 10

Sample Output 2:
-0.25+i2.22
-0.25-i2.22, I have written this Solution Code: import java.io.*;
import java.util.*;



class Main {
	public static void main (String[] args) throws IOException {

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String s=br.readLine().toString();
		String[] str = s.split(" ");
		float a=Float.parseFloat(str[0]);
		float b=Float.parseFloat(str[1]);
		float c=Float.parseFloat(str[2]);
		float div = (float)(b*b-4*a*c);
		if(div>0.0){
			float alpha= (-b+(float)Math.sqrt(div))/(2*a);
			float beta=  (-b-(float)Math.sqrt(div))/(2*a);
			System.out.printf("%.2f\n",alpha);
			System.out.printf("%.2f",beta);
		}
		else{
			float rp=-b/(2*a);
			float ip=(float)Math.sqrt(-div)/(2*a);
			System.out.printf("%.2f+i%.2f\n",rp,ip);
			System.out.printf("%.2f-i%.2f\n",rp,ip);
		}


	}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?