C++求一元四次方程的存在根

Question

材料内容是：
给定实数 a, b, c 。 研究双二次方程 ax4 + bx2 + c = 0，即找到给定方程的所有实根并表示出来，如果没有请写出
我写了一半发现我的代码不对劲，改了三个小时，不清楚是哪的问题

cout << "ax^4+bx^2+c=0" << endl;
    float x1,x2,x3,x4, a, b, c, discriminant;
    cout << "a:"; cin >> a;
    cout << endl;
    cout << "b:"; cin >> b;
    cout << endl;
    cout << "c:"; cin >> c;
    cout << endl;
    discriminant = b * b - 4 * a * c;
    x1 = sqrt((-b + sqrt(discriminant)) / (2 * a));
    x2=-sqrt((-b + sqrt(discriminant)) / (2 * a));
    x3= sqrt((-b - sqrt(discriminant)) / (2 * a));
    x4=-sqrt((-b - sqrt(discriminant)) / (2 * a));
    cout << "x1=" << x1 << endl;
    cout << "x2=" << x2 << endl;
    cout << "x3=" << x3 << endl;
    cout << "x4=" << x4 << endl;

Answer 1

#include <iostream>

int main()
{
    double a, b, c, d;
    std::cin >> a >> b >> c;
    d = b * b - 4 * a * c;
    if (d < 0)
    {
        std::cout << "no real solution\n";
        return 0;
    }
    d = std::sqrt(d);
    int n = 0;
    if (-b - d >= 0)
    {
        double x = std::sqrt((-b - d) / (2 * a));
        std::cout << x << " " << -x << " ";
        n += 2;
    }
    if (-b + d >= 0)
    {
        double x = std::sqrt((-b + d) / (2 * a));
        std::cout << x << " " << -x << " ";
        n += 2;
    }
    if (n == 0)
        std::cout << "no real solution\n";
    return 0;
}