Quadratic equations, need some help

[DaRok28]

// Program Description:
// This program solves quadratic equations to find their roots. This
// program takes values of a, b, and c as input and outputs the root(s).
// The user can repeat the calculation for as many equations as they like.

#include <iostream>
#include <cmath>
#include <complex>
using namespace std;
 
int main()
{
   char cAgain;
   int iDataset = 1;
   double dRoot1, dRoot2, dValuea, dValueb, dValuec, dDescriminant;
 
   cout.setf(ios::fixed);
   cout.setf(ios::showpoint);
   cout.precision(4);
 
   do
   {
      cout << endl << "****************" << endl
           << "* Data Set " << iDataset << " *" << endl
           << "****************" << endl << endl;
      cout << endl << "Please enter coefficients for quadratic equation"
           << " ax^2 + bx + c = 0" << endl;
      cout << endl << "Value of 'a' : ";
      cin >> dValuea;
      cout << "Value of 'b' : ";
      cin >> dValueb;
      cout << "Value of 'c' : ";
      cin >> dValuec;
      dRoot1 = (- dValueb + sqrt(pow(dValueb, 2) - 4 * dValuea * dValuec)) / 2 * dValuea;
      dRoot2 = (- dValueb - sqrt(pow(dValueb, 2) - 4 * dValuea * dValuec)) / 2 * dValuea;
      dDescriminant = pow(dValueb, 2) - 4 * dValuea * dValuec;
 
         if(dDescriminant < 0)
         {
            cout << endl << "You have entered the equation " << dValuea
                 << "x^2 + " << dValueb << "x +" << dValuec
                 << " = o" << endl;
            cout << "The root to this equation is: " << endl << endl
                 << dRoot1 << " i  and" << endl << dRoot2 << endl << endl;
 
 
         }
         else if(dDescriminant == 0)
         {
            cout << endl << "You have entered the equation " << dValuea
                 << "x^2 + " << dValueb << "x +" << dValuec
                 << " = 0" << endl;
            cout << "The root to this equation is: " << endl << endl
                 << dRoot1 << endl;
         }
         else
         {
            cout << endl << "You have entered the equations " << dValuea
                 << "x^2 + " << dValueb << "x +" << dValuec
                 << " = o" <<endl;
            cout << "The root to this equations is: " << endl << endl
                 << dRoot1 << " and" << endl << endl << dRoot2 << endl;
         }
      cout << endl << "Want to do this again? Y/y/N/n" << endl << endl;
      cin >> cAgain;
      iDataset++;
   }while (cAgain == 'Y' || cAgain == 'y');
 
   return 0;
 
}
 

My else if and else statements both work the way I want them too, but I need my if tatement to do negative numbers. When I enter the values for a, b, c as
1, 4, a nd 7 respectfully I need the answer to be -2.0000 - 1.7321 i and -2.0000 + 1.7321 i. I am just not sure how to do this, so if anyone has any suggestions.

[horace1]

you need to test

     dDescriminant = pow(dValueb, 2) - 4 * dValuea * dValuec;
         if(dDescriminant < 0)
         {
 

before you calculate dRoot1 and dRoot2

if dDescriminant>0 is real rools (note your calculation is not quite correct)

      dRoot1 = (- dValueb + sqrt(pow(dValueb, 2) - 4 * dValuea * dValuec)) / 2 * dValuea;
      dRoot2 = (- dValueb - sqrt(pow(dValueb, 2) - 4 * dValuea * dValuec)) / 2 * dValuea;
 

if dDescriminant< 0
the real part is

        - dValueb /( 2 * dValuea)
 

the imaginary parts are

   + and -   sqrt(-dDescriminant) / (2 * dValuea)
 

[DaRok28]

Thanks, I am not sure if what I did is correct, but i went off of what you told me and it worked, so thanks

you need to test

     dDescriminant = pow(dValueb, 2) - 4 * dValuea * dValuec;
         if(dDescriminant < 0)
         {
 

before you calculate dRoot1 and dRoot2

if dDescriminant>0 is real rools (note your calculation is not quite correct)

      dRoot1 = (- dValueb + sqrt(pow(dValueb, 2) - 4 * dValuea * dValuec)) / 2 * dValuea;
      dRoot2 = (- dValueb - sqrt(pow(dValueb, 2) - 4 * dValuea * dValuec)) / 2 * dValuea;
 

if dDescriminant< 0
the real part is

        - dValueb /( 2 * dValuea)
 

the imaginary parts are

   + and -   sqrt(-dDescriminant) / (2 * dValuea)