一元二次方程的求解

大佬们这个一元二次方程的求解方法哪里有什么问题吗？
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main(int argc, char** argv) {
	float a,b,c;
	cin>>a>>b>>c;
	if(b*b==4*a*c){
		//cout<<"x1=x2="<<fixed<<setprecision(5)<<-b/(2*a);
		printf("x1=x2=%0.5f",-b/(2*a));
		return 0;
	}
	if(b*b>4*a*c){
		//cout<<"x1="<<fixed<<setprecision(5)<<(-b+sqrt(b*b-4*a*c))/(2*a)<<";"<<"x2="<<fixed<<setprecision(5)<<(-b-sqrt(b*b-4*a*c))/(2*a);
		printf("x1=%0.5f;x2=%0.5f",(-b+sqrt(b*b-4*a*c))/(2*a),(-b-sqrt(b*b-4*a*c))/(2*a));
		return 0;
	}
	if(b*b<4*a*c){
		//cout<<"x1="<<fixed<<setprecision(5)<<(-b/(2*a))<<"+"<<fixed<<setprecision(5)<<(sqrt(4*a*c-b*b)/(2*a))<<"i"<<";"<<"x2="<<fixed<<setprecision(5)<<(-b/(2*a))<<"-"<<fixed<<setprecision(5)<<(sqrt(4*a*c-b*b)/(2*a))<<"i";
		printf("x1=%0.5f+%0.5fi;x2=%0.5f-%0.5fi",-b/(2*a),(sqrt(4*a*c-b*b)/(2*a)),-b/(2*a),(sqrt(4*a*c-b*b)/(2*a)));
		return 0;
	}
	return 0;
}