编写一个程序求一元二次方程ax bx

PrivateSubCommand1_Click()DimaAsSingle,bAsSingle,cAsSingleDimdAsSingle,x1AsSingle,x2AsSinglea=InputB

什么程序?c语言的?说下想法吧先求△=b²-4ac之后分类判断（1）△0,x=(-b±根号（b²-4ac）)/(2a)再问：能不能写成IF……ELSE……end……的形式再答：de

double x1 = 0;//解1double x2 = 0;//解2Console.WriteLine("求 ax^

//Equation.h#ifndef_Equation_h#define_Equation_hclassEquation{private:doublea;doubleb;doublec;voidSh

#include"stdio.h"#include"math.h"doublex1,x2,p;floatfile1(floata,floatb){x1=(-b+sqrt(p))/2*a;x2=(-b-

PublicClassForm1PrivateSubButton1_Click(ByValsenderAsSystem.Object,ByValeAsSystem.EventArgs)HandlesB

vara,b,c,m:real;beginreadln(a,b,c);m:=b*b-4*a*c;ifm>0thenbeginwrite((-1*b+sqrt(m))/(2*a):0:3);write(

这个可以这样做~cleara=rand(10,1);%产生一组随机数b=rand(10,1);%产生另一组随机数b=b-(dot(a,b)/dot(a,a)).*a;%可以使用施密特正交化的方法转化d

请参考

usingSystem;usingSystem.Collections.Generic;usingSystem.Text;usingSystem.Collections;namespacecacFC{

#include#includevoidmain(){floata,b,c,disc,x1,x2,realpart,imagpart;scanf("%f,%f,%f",&a,&b,&c);/*以a,b

#include"stdio.h"#include"math.h"voidmain(){floata,b,c;floatdelta;printf("inputa:");scanf("%f",&a);p

a=-10;b=10;n=0;whileb-a>epst=(a+b)/2;n=n+1;if4*t^2+3*t-6==0break;elseif(4*a^2+3*a-6)*(4*t^2+3*t-6)>0

第二题：#includevoidmain(){inti,g,s,b;for(i=100;i

C++的代码：#include#includevoidmain(void){doublea,b,c,d;charch('y');do{coutb>>c;if(-0.0001

#include#include

dimaasdouble,basdouble,casdoubledimx1asdouble,x2asdoublea=val(inputbox(""))b=val(inputbox(""))c=val(

PrivateSubCommand1_Click()Dima#,b#,c#,d#,x1#,x2#a=Val(InputBox("a=","数据输入框",1))b=Val(InputBox("b=","

C++的代码：#include<iostream.h>#include<math.h>voidmain(void){doublea,b,c,d;charch('y');do{c

C++的代码：#include#includevoidmain(void){doublea,b,c,d;charch('y');do{coutb>>c;if(-0.0001