C语言解X3+AX2+B=0 一元三次的程序main(){Float l,g1=0.03275,f,g2=0.06544,A,B,t1=-5,t2=-5,c=68.51,x,x1,x0,dx;scanf("%f",&l);A=c-(73000*g1*g1*l*l)/(24*c*c)-73000*0.0000196*(t2-t1);B=73000*g2*g2*l*l/24;printf("%f,%f",A,B);x0=1.0;do{dx=(x0*x0*x0-

C语言解X3+AX2+B=0 一元三次的程序main(){Float l,g1=0.03275,f,g2=0.06544,A,B,t1=-5,t2=-5,c=68.51,x,x1,x0,dx;scanf("%f",&l);A=c-(73000*g1*g1*l*l)/(24*c*c)-73000*0.0000196*(t2-t1);B=73000*g2*g2*l*l/24;printf("%f,%f",A,B);x0=1.0;do{dx=(x0*x0*x0-
C语言解X3+AX2+B=0 一元三次的程序
main()
{
Float l,g1=0.03275,f,g2=0.06544,A,B,t1=-5,t2=-5,c=68.51,x,x1,x0,dx;
scanf("%f",&l);
A=c-(73000*g1*g1*l*l)/(24*c*c)-73000*0.0000196*(t2-t1);
B=73000*g2*g2*l*l/24;
printf("%f,%f",A,B);
x0=1.0;
do{
dx=(x0*x0*x0-A*x0*x0-B)/(3*x0*x0-2*A*x0);
x1=x0;
x0=x0-dx;
}
while(x0!=x1);
printf("\nc=%f\n",x0);
f=(g2*l*l)/(8*x0);
printf("f=%f",f);
}
问题出在哪里,求改正,写出正确的代码

C语言解X3+AX2+B=0 一元三次的程序main(){Float l,g1=0.03275,f,g2=0.06544,A,B,t1=-5,t2=-5,c=68.51,x,x1,x0,dx;scanf("%f",&l);A=c-(73000*g1*g1*l*l)/(24*c*c)-73000*0.0000196*(t2-t1);B=73000*g2*g2*l*l/24;printf("%f,%f",A,B);x0=1.0;do{dx=(x0*x0*x0-
X^3+A*X^2+B=0?是这意思吗?