（２次方程式を解くサンプル実行例）

> java qequ 1 5 6
1.0x^2 + 5.0x + 6.0 = 0   : x = -3.0 , -2.0 (実数解)

> java qequ 1 6 9
1.0x^2 + 6.0x + 9.0 = 0   : x = -3.0 (重解)

> java qequ 1 3 6
1.0x^2 + 3.0x + 6.0 = 0   : x = -1.5±1.9364917i (虚数解)


（以下はコンパイラに処理させたソースコード）


Sub main(argv[] as string)
  dim a,b,c as float
  
  
  //引数の数をチェック
  if (int)argv<3 then
    call println("usage : java qequ a b c")
    exit sub
  endif

  //係数
  a = argv[0]
  b = argv[1]
  c = argv[2]

  //2次方程式なのでa!=0でないとこまる
  if a = 0 then
    call println("err a = 0")
    exit sub
  endif

  Call ResolveQuadricEquation(a,b,c)

end sub

//２次方程式の解を求める
Sub ResolveQuadricEquation(a as float,b as float,c as float)
  dim d , r as float
  dim x[2] as float
  dim st as string
  //判別式
  d = b*b - 4 * a * c
  
  //解の表示のための文字列
  st = a + "x^2 + " + b + "x + " + c + " = 0   : x = "
  call print(st)
  
  if d > 0.0001 then
    //実数解が2つ
    r = sqr(d)
    
    x[0] = (-b-r) / (2 * a)
    x[1] = (-b+r) / (2 * a)
    call println(x[0] + " , " +x[1] + " (実数解)")

  elseif d < -0.0001 then
    //虚数解が2つ
    r = sqr(-d)
    
    x[0] = -b / (2 * a)
    x[1] = r / (2 * a)
    call println(x[0] + "±" +x[1] + "i (虚数解)")

  else
    //解は１つだけ
    x[0] = -b / (2 * a)
    call println(x[0] + " (重解)")

  endif
  
end sub

//平方根をニュートン法で大雑把に求める
//RTClassを経由してMath.powを呼んでもよいが、せっかくなので自前でやってみる
Function sqr(f as float) as float
  dim i as int
  dim x as float

  //x' = (x^2 + f) / (2 * x) を反復計算
  x = f
  for i = 0 to 5
    x = (x * x + f) / (2 * x)
  next
  
  return x
  
end function