Fist Post, multiple questions ;)

[Deimos]

Hello everyone,
I am a 15 year old trying to lean programing by myself, I have read some books and online tutorials and I have several questions for you experienced programmers out there. :D

1st question:
what exactly is a(the?) buffer? From my readings it seems to be some sort of intermediary memory that temporarily stores data from and to the console output but I'm not sure.

2nd question
Is there any difference from the string name type and the char string[LENGTH]?

3rd question
what can the .put() and get() do that << and >> can't?

4th question
Its a work for school I need to make a program that can calculate the result of a second degree equation, here is the code, so far:

#include<iostream>
#include<cmath>
#include<cstdlib>

using namespace std;

int main(){
    float a, b, c, x1, x2, sq;
    char sn;
    cout << "\t** A formula e' x = (-b +/- Sqrt(b^2 - 4ac))/2a **" << endl;//I'm Portuguese so this is a Portuguese string, it names the base equation
    do{
       cout << "Inserir 'a' -> ";cin >> a;//asks for a, b and c
       cout << "Inserir 'b' -> ";cin >> b;
       cout << "Inserir 'c' -> ";cin >> c;
       sq = b*b - 4*a*c;//Calculates a chunk of the expression for possible/impossible test
       if(sq >= 0){//if it isn't going to give us the sqrt of a negative number...
             x1 = ((b*-1) + sqrt(sq))/2*a;//calculate both solutions
             x2 = ((b*-1) - sqrt(sq))/2*a;
             cout << "\nx = " << x1 << " V x = " << x2 << endl;//give out the solution
             }else{//inform the user of the fact that the values he inserted lead to an impossible equation
             cout << "\nEquacao Impossivel" << endl;
             }
       cout << "Deseja inserir outro valor?(S/N): ";cin >> sn;//asks for a second set of values
       }while(toupper(sn) == 'S');
       system("Pause");
}
    

it gets some right and some wrong, for example:
a-> -1;
b-> 4;
c-> -4;

x = 2 V x = 2 CORRECT!!

a-> 6;
b-> -1;
c-> -1;

x = 18 V x = -12 WRONG!! x = 1/2(0.5) V x = 1/3(0.33333...)

Best Regards and thank you in advance

Deimos

[psault]

For your problem, change your equations:

x1 = ((b*-1) + sqrt(sq))/(2*a);
x2 = ((b*-1) - sqrt(sq))/(2*a);

[Deimos]

I'm sorry, I'm kinda slow, whats wrong with them?

[psault]

Sometimes you need to help the compiler specify the order of operations for doing math. Notice that I added parenthesis around your denominator values so that the compiler does not perform the division before the multiplication.