Quadratic Equations Solver

[B+]

Is this homework?

 Try x1 = (-b + SQR(b^2 - 4*a*c))/(2*a) and  x2 = (-b + SQR(b^2 - 4*a*c))/(2*a)

But you better check D =  b^2 - 4*a*c > 0 first because SQR doesn't do negative SQR's because that needs imaginary numbers.

[ntech]

It's straightforward if you hard code it, but to parse a string containing a quadratic equation is hard.

[tenochtitlanuk]

You might usefully look at
rosettacode.org/wiki/Roots_of_a_quadratic_function
.....which looks at this in many languages and discusses some of the problem cases.

[tsh73]

but to parse a string containing a quadratic equation is hard

but to create GUI with freeform with boxes for 'a', 'b' and 'c' is easy



'   Form created with the help of Freeform-J v.261006
'   Generated on Mar 11, 2020 at 23:05:29

   nomainwin

   WindowWidth = 520
   WindowHeight = 405

   UpperLeftX=int((DisplayWidth-WindowWidth)/2)
   UpperLeftY=int((DisplayHeight-WindowHeight)/2)

   statictext #main.statictext1, "a", 26, 16, 25, 20
   statictext #main.statictext2, "b", 134, 16, 32, 20
   statictext #main.statictext3, "c", 246, 16, 32, 25
   textbox #main.a, 22, 46, 100, 25
   textbox #main.b, 134, 46, 100, 25
   textbox #main.c, 246, 46, 100, 25
   button #main.button7, "Solve!", [button7Click], UL, 366, 21, 120, 45
   texteditor #main.log, 22, 86, 472, 255
   menu #main, "Edit"    '<--- Texteditor Menu can be moved but not removed.

   open "Quadratic equation solver" for window_nf as #main
   print #main, "trapclose [quit.main]"

   print #main, "font ms_sans_serif 10"

   print #main.a, 1
   print #main.b, 0
   print #main.c, -1

   wait

[quit.main]
   Close #main
   END


[button7Click]    'Perform action for the button named 'button7'
   'Insert your own code here
   #main.a "!contents? a"
   #main.b "!contents? b"
   #main.c "!contents? c"

   print #main.log, "";a;"*x^2";"+";b;"*x";"+";c;"=0"
   print #main.log,"you solve it"
   wait


[ice]

Mar 11, 2020 17:42:59 GMT B+ said:

Is this homework?

 Try x1 = (-b + SQR(b^2 - 4*a*c))/(2*a) and  x2 = (-b + SQR(b^2 - 4*a*c))/(2*a)

But you better check D =  b^2 - 4*a*c > 0 first because SQR doesn't do negative SQR's because that needs imaginary numbers.

Yes, it was homework.

[ice]

Thank You. This helped a lot.

[B+]

Oops x2 above is same and x1, sorry I copied x1 to make x2 but forgot to replace the pos sign with neg sign right after -b.

Should be x2 = (-b - SQR(b^2 - 4*a*c))/(2*a)

Hey looks like tsh73 has all the hard work done for you getting setup on JB, just plug in formulas and program it to write answers in the Text Editor.

Man I wish I had JB when I was taking math classes :-))