Solving the quadratic equation, but getting weird errors

Question

I am trying to execute my first Fortran program trying to solve a quadratic equation. I double and triple checked my code and see nothing wrong. I keep getting "Invalid character in name (1)" and "Unclassified statement in (1)" in different places. What is wrong with the code?

! This program solves quadratic equations
! of the form ax^2 + bx + c = 0.
! Record:
! Name:             Date:     Notes:
! Damon Robles      4/3/10    Original Code

PROGRAM quad_solv
IMPLICIT NONE

! Variables
REAL :: a, b, c
REAL :: discrim, root1, root2,    
COMPLEX :: comp1, comp2
CHARACTER(len=1) :: correct 

! Prompt user for coefficients.
WRITE(*,*) "This program solves quadratic equations "
WRITE(*,*) "of the form ax^2 + bx + c = 0. "
WRITE(*,*) "Please enter the coefficients a, b, and "
WRITE(*,*) "c, separated by commas:"
READ(*,*) a, b, c
WRITE(*,*) "Is this correct: a = ", a, " b = ", b
WRITE(*,*) " c = ", c, " [Y/N]? "
READ(*,*) correct
IF correct = N STOP
IF correct = Y THEN

! Definition
discrim = b**2 - 4*a*c

! Calculations
IF discrim > 0 THEN
root1 = (-b + sqrt(discrim))/(2*a)
root2 = (-b - sqrt(discrim))/(2*a)
WRITE(*,*) "This equation has two real roots. "
WRITE(*,*) "x1 = ", root1
WRITE(*,*) "x2 = ", root2
IF discrim = 0 THEN
root1 = -b/(2*a)
WRITE(*,*) "This equation has a double root. "
WRITE(*,*) "x1 = ", root1
IF discrim < 0 THEN
comp1 = (-b + sqrt(discrim))/(2*a)
comp2 = (-b - sqrt(discrim))/(2*a)
WRITE(*,*) "x1 = ", comp1
WRITE(*,*) "x2 = ", comp2

PROGRAM END quad_solv

Answer 1

, - , Fortran >= 90. , -, - , IF, , , , , . (, "A" , .) iostat , IF , DO , .

program test_readdata

real :: a, b, c
integer :: ReturnCode
character (len=1) :: correct

ReadData: do

   write (*, '( "This program solves quadratic equations of the form ax^2 + bx + c = 0. " )' ) 
   write (*, '( "Please enter the coefficients a, b, and c, separated by commas: " )', advance='no' )
   read (*,*, iostat=ReturnCode) a, b, c
   if ( ReturnCode /= 0 ) cycle ReadData
   write (*,*) "Is this correct: a = ", a, " b = ", b, " c = ", c
   write (*, '( "Enter Y or N: " )', advance='no' )
   read (*,*, iostat=ReturnCode) correct
   if ( ReturnCode /= 0 ) cycle ReadData
   if ( correct == 'Y'  .or.  correct == 'y' )  exit ReadData

end do ReadData

stop

end program test_readdata

: Fortran 95/2003 , .

Answer 2

, , - .

END

?

ELF 90 , . Lahey ELF 95 Visual Studio , .

Answer 3

... , ...

! This program solves quadratic equations
! of the form ax^2 + bx + c = 0.
! Record:
! Name:             Date:     Notes:
! Damon Robles      4/3/10    Original Code

PROGRAM quad_solv

  IMPLICIT NONE

  ! Variables
  REAL :: a, b, c
  REAL :: discrim, root1, root2    
  COMPLEX :: comp1, comp2
  CHARACTER(len=1) :: correct 

  ! Prompt user for coefficients.
  WRITE(*,*) "This program solves quadratic equations "
  WRITE(*,*) "of the form ax^2 + bx + c = 0. "
  WRITE(*,*) "Please enter the coefficients a, b, and "
  WRITE(*,*) "c, separated by commas:"
  READ(*,*) a, b, c
  WRITE(*,*) "Is this correct: a = ", a, " b = ", b
  WRITE(*,*) " c = ", c, " [Y/N]? "
  READ(*,*) correct

  IF (correct == 'N') STOP

  IF (correct == 'Y') THEN

    ! Definition
    discrim = b**2 - 4*a*c

    ! Calculations
    IF (discrim > 0) THEN
      root1 = (-b + sqrt(discrim))/(2*a)
      root2 = (-b - sqrt(discrim))/(2*a)
      WRITE(*,*) "This equation has two real roots. "
      WRITE(*,*) "x1 = ", root1
      WRITE(*,*) "x2 = ", root2
    ELSEIF (discrim == 0) THEN
      root1 = -b/(2*a)
      WRITE(*,*) "This equation has a double root. "
      WRITE(*,*) "x1 = ", root1
    ELSE
      comp1 = (-b + sqrt(discrim))/(2*a)
      comp2 = (-b - sqrt(discrim))/(2*a)
      WRITE(*,*) "x1 = ", comp1
      WRITE(*,*) "x2 = ", comp2
    END IF

  END IF  

END PROGRAM quad_solv

P.S. . P.P.S. , STOP. ( ) . STOP - END PROGRAM, . STOP .

Answer 4

Some changes: correct IF syntax; constants must be real numbers, not integers, if necessary:

IF (CORRECT == "N" ) stop
if ( discrim > 0.0 ) then

Apply in extra places and you have to be very close. Good luck.

It might not be a good algorithm to check for a floating point number to get an exact value. What if the rounding error makes disrim value 1.0E-30, when perfect arithmetic gives a value of 0 and indicates a double root?