Question

编程，求方程2X²+3X-3=0的根.

Answer 1

这是求任意一元二次方程解的代码，把它改成你那个应该很简单了

#include <iostream>

#include <cmath>

using namespace std;

int main(){

double a, b, c, x1, x2,d;

cout << "输入a,b,c"<<endl;

cin >> a >> b >> c;

d = b*b - 4 * a*c;

if (a == 0){

if (b == 0 && c == 0)

cout << "这是一个恒等式" << endl;

else if (b == 0 && c != 0)

cout << "方程无解" << endl;

else

{

x1 = -c / b;

cout << "方程只有一个解，x= " << x1 << endl;

}

}

else if (d==0)

{

x1 = -b / (2 * a);

cout << "方程有一对相同的解，x1=x2= " << x1;

}

else if (d<0)

{

cout << "方程无解" << endl;

}

else

{

x1 = (-b + sqrt(d)) / (2 * a);

x2 = (-b - sqrt(d)) / (2 * a);

cout << "方程有两个实根，x1=" << x1 << ",x2=" << x2 << endl;

}

return 0;

}

Answer 2

public class Test1 {
public static void main(String[] args) {
int a = 2;
int b = 3;
int c = -3;

double answer1,answer2;
answer1 = (-b+Math.sqrt((Math.pow(b, 2)-4*a*c)))/2*a;
answer2 = (-b+Math.sqrt((Math.pow(b, 2)-4*a*c)))/2*a;
System.out.println(answer1);
System.out.println(answer2);
}
}