## Algorithm

Problem Name: beecrowd | 1020

Adapted by Neilor Tonin, URI Brazil

Timelimit: 1

Read 3 floating-point numbers. After, print the roots of bhaskara’s formula. If it's impossible to calculate the roots because a division by zero or a square root of a negative number, presents the message “Impossivel calcular”.

## Input

Read 3 floating-point numbers (double) A, B and C.

## Output

Print the result with 5 digits after the decimal point or the message if it is impossible to calculate.

 Input Samples Output Samples 10.0 20.1 5.1 R1 = -0.29788 R2 = -1.71212

 0.0 20.0 5.0 Impossivel calcular

 10.3 203.0 5.0 R1 = -0.02466 R2 = -19.68408

 10.0 3.0 5.0 Impossivel calcular

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````

#include <bits/stdc++.h>
#include <math.h>
using namespace std;

int main(){
double a,b,c,t,R1,R2;
cin>>a>>b>>c;

if(((b*b)-4*a*c)<0 ||a==0){
cout<<"Impossivel calcular"<<endl;
}else{
t=sqrt((b*b)-4*a*c);
R1=((-b + t) / (2 * a));
R2=((-b - t) / (2 * a));
cout<<fixed;
cout<<setprecision(5)<<"R1 = "<<R1<<endl;
cout<<setprecision(5><<"R2 = "<<R2<<endl;
}
return 0;
}
``````
Copy The Code &

Input

cmd
10.0 20.1 5.1

Output

cmd
R1 = -0.29788 R2 = -1.71212

### #2 Code Example with C++ Programming

```Code - C++ Programming```

``````

#include <stdio.h>
#include <math.h>

int main()
{
double a, b, c, t;
scanf("%lf %lf %lf", &a, &b, &c);

if(((b * b) - 4 * a * c) < 0 || a == 0) {
printf("Impossivel calcular\n");
} else {
t = sqrt((b * b) - 4 * a * C);
printf("R1 = %.5lf\nR2 = %.5lf\n", ((-b + t) / (2 * a)), ((-b - t) / (2 * a))>;
}

return 0;
}
``````
Copy The Code &

Input

cmd
0.0 20.0 5.0

Output

cmd
Impossivel calcular

### #3 Code Example with Java Programming

```Code - Java Programming```

``````

import java.util.Scanner;

public class Main {
public static void main(String[] args) {
double A, B, C, R1, R2;
Scanner input = new Scanner(System.in);
A = input.nextDouble();
B = input.nextDouble();
C = input.nextDouble();

if ((A == 0) || (((B * B) - (4 * A * C))  <  0)) {
System.out.print("Impossivel calcular\n");
} else {
R1 = ((-B + Math.sqrt(((B*B) -(4*A*C)))) / (2*A));
R2 = ((-B - Math.sqrt(((B*B) -(4*A*C)))) / (2*A));

System.out.printf("R1 = %.5f\n", R1);
System.out.printf("R2 = %.5f\n", R2);

}
}
}
``````
Copy The Code &

Input

cmd
10.3 203.0 5.0

Output

cmd
R1 = -0.02466 R2 = -19.68408

### #4 Code Example with Python Programming

```Code - Python Programming```

``````

import math
A,B,C = map(float,input().split())
D = (B**2)-(4*A*C)
if(D <0 or A==0):
print("Impossivel calcular")
else:
D=math.sqrt(D)
R1 = (-B+D)/(2*A)
R2 = (-B-D)/(2*A)
print(f'R1 = {R1:0.5f}\nR2 = {R2:0.5f}'>
``````
Copy The Code &

Input

cmd
10.0 20.1 5.1

Output

cmd
R1 = -0.29788 R2 = -1.71212