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: /**
 *      Author : tourist1256
 *      Time   : 2022-01-19 02:44:22
 **/

#include <bits/stdc++.h>

using namespace std;

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif

int main() {
    float a, b, c;
    float root1, root2, imaginary;
    float discriminant;
    scanf("%f%f%f", &a, &b, &c);
    discriminant = (b * b) - (4 * a * c);
    switch (discriminant > 0) {
        case 1:
            root1 = (-b + sqrt(discriminant)) / (2 * a);
            root2 = (-b - sqrt(discriminant)) / (2 * a);

            printf("%.2f\n%.2f", root1, root2);
            break;

        case 0:
            switch (discriminant < 0) {
                case 1:
                    root1 = root2 = -b / (2 * a);
                    imaginary = sqrt(-discriminant) / (2 * a);

                    printf("%.2f + i%.2f\n%.2f - i%.2f", root1, imaginary, root2, imaginary);
                    break;

                case 0:
                    root1 = root2 = -b / (2 * a);

                    printf("%.2f\n%.2f", root1, root2);

                    break;
            }
    }
    return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?