two errors which I don't know how to fix

[davidoffc]

Hi all. On "cout << "x1 = " << x1= ((-b) - sqrt(D)/ (2*a)) << endl; cout << "x2 = " << x2= ((-b) + sqrt(D)/ (2*a)) << endl;}" lines i get errors:
invalid operands of types 'float' and '<unresolved overloaded function type>'....
How to fix them?
Thanks

  #include <iostream>
#include <cmath>
using namespace std;

int main(void)
{
    int a, b, c;
    float  x1, x2, x, D;
cout << "Iveskite koeficienta a: ";
cin >> a;
cout << "Iveskite koeficienta b: ";
cin >> b;
cout << "Iveskite koeficienta c: ";
cin >> c;

if (a != 0)
{
     {D=(b*b)-(4*a*c);}
     if (D>=0)
     {
     cout << "x1 = " << x1= ((-b) - sqrt(D)/ (2*a)) << endl;
     cout << "x2 = " << x2= ((-b) + sqrt(D)/ (2*a)) << endl;}

     else
     cout << "Diskriminantas neigiamas. Sprendiniu nera." << endl;}



else          if ((a == 0))
                cout << "x = " << (x = -c/b ) << endl;
                else
                cout << "Irasyk normalius skaicius, tau cia ne cirkas." <<endl;

return 0;
}

[Null]

First calculate x1 and x2, then print the results

   
// ...
     x1= ((-b) - sqrt(D)/ (2*a));
     x2= ((-b) + sqrt(D)/ (2*a));

     cout << "x1 = " << x1 << endl;
     cout << "x2 = " << x2 << endl;

[Peter87]

The problem is that = has higher precedence than << so the compiler will see it as
(cout << "x1 = " << x1) = (((-b) - sqrt(D)/ (2*a)) << endl);

You can use parentheses to fix it
cout << "x1 = " << (x1= ((-b) - sqrt(D)/ (2*a))) << endl;

Putting the assignment of x1 and x2 on separate lines, as Null said, is probably easier and makes the code easier to understand.

[davidoffc]

Thanks :)

[davidoffc]

By the way, this program cannot calculate correctly... Maybe you could see the problem. The main thing is that when i write, for example, a=1, b = -5, c=6, this program calculate x1= 4,5, x2 = 5,5, but x1 and x2 should be 2 and 3. What's the problem? :/

[Peter87]

-b will not be divided. Use parentheses to make it work.

[davidoffc]

nah, Peter, -b works good. The main problem was, that firstly i should make
((-b) +- sqrt(D)) and only then /2*a . ;]

[Chervil]

That's exactly what Peter meant. He said "will not be divided".

Also, these parentheses serve no purpose, but add to the clutter: (-b)
all you need is simply
(-b - sqrt(D)) / (2*a)