Hey steemit and steemians.

Today is kinda best day I wanna share the best program for those who dont know it, lets get into it.

ProgramBody:

#include <iostream>

#include <string>

using namespace std;

class eq{

private:

    double a,vi,vf,s,t,vi1,vf2,v;

public:

    void solve(){

    cout<<"                    PHENIOX 1.0"<<endl;

    cout<<endl;

    cout<<endl;

    cout<<"            EXEUTION HELPER(Compulsory)"<<endl;

    cout<<"             Enter the given data as well as required data."<<endl;

    cout<<"             The value you want to find put its value to 101.11 ."<<endl;

    cout<<"             The values who arent in this equation should be entered as 101.12(VI)."<<endl;

    cout<<"             Some times values are given in square form so be careful while reading answer."<<endl;

    cout<<endl;

    cout<<" WARNING!! this version is basic one it will only work with single value finding."<<endl;

    cout<<endl;

    cout<<"                    LETS GET STARTED WITH PHENIOX 1.0"<<endl;

    cout<<endl;

    cout<<endl;

    cout<<"Enter The Value Of a"<<endl;

    cin>>a;

    cout<<"Enter The Value Of vi"<<endl;

    cin>>vi;

    cout<<"Enter The value of vf"<<endl;

    cin>>vf;

    cout<<"Enter The Value Of s"<<endl;

    cin>>s;

    cout<<"Enter The Value Of t"<<endl;

    cin>>t;

    cout<<endl;

    vi1=(vi*vi);

        vf2=(vf*vf);

    if(a==101.11){

        if(s==101.12){

            cout<<"value of a is "<<(vf-vi)/t<<endl;

        }else if(t==101.12){

        cout<<"value of a is "<<(vf2-vi1)/(2*s)<<endl;

        }

    }

    if(vi==101.11){

        if(s==101.12){

            cout<<"value of vi is "<<vf/(a*t)<<endl;

        }else if(t==101.12){

            double c,d;

            c=((2*a*s)/vf2);

            d=-c;

        cout<<"value of vi is "<<d<<endl;

        }

    }

    if(vf==101.11){

    if(s==101.12){

    cout<<"Value of vf is "<<vi+(a*t)<<endl;

    }else if(t==101.12){

    cout<<"Value of vf is "<<(2*a*s)+vi1<<endl;

    }

    }

    if(s==101.11){

            cout<<"Enter The value of V(if any if not put it as 101.12)."<<endl;

            cin>>v;

        if(t==101.12&&v==101.12){

            cout<<"Value Of s is "<<(vf2-vi1)/(2*a)<<endl;

        }else if(vi==101.12&&a==101.12){

        if(vf==101.12){

            cout<<"value of s is"<<v*t<<endl;

        }

        }

    }

    if(t==101.11){

            cout<<"Enter The value of V(if any if not put it as 101.12)."<<endl;

    cin>>v;

        if(s==101.12&&v==101.12){

            cout<<"value of t is"<<(vf-vi)/a<<endl;

        }else if(a==101.12&&vi==101.12){

        if(vf==101.12){

            cout<<"value of t is"<<s/t<<endl;

        }

        }

    }

    }

};

int main (){

eq me;

me.solve();

}

this program is best I made, about year ago Now I was busy so I did't make more but I will.

Executation:

Here is startup of program in code blocks. 

Place or input the values according to given instructions here.

Answer will be displayed here, dont worry about this answer beacuse the vlaues inputed are super strange you can do it for your self, for real values you will get correct answer.

THANK YOU ALL