This program graphs the equation X^3/3 - 2*X + 5. When you click on the graph, it uses Newton's method to find a root of the equation, starting from the X value that you clicked.

' The function.
Private Function F(ByVal X As Double) As Single
    F = X * X * X / 3 - 2 * X * X + 5
End Function

' Draw the background graph.
Private Sub DrawGraph()
Const STEP_SIZE As Single = 0.1
Dim X As Single
Dim Y As Single
Dim y1 As Single
Dim y2 As Single

    Me.Cls

    ' Draw the axes.
    For X = -10 To 10
        Me.Line (X, -0.5)-Step(0, 1), vbBlue
    Next X
    For Y = -10 To 10
        Me.Line (-0.5, Y)-Step(1, 0), vbBlue
    Next Y
    Me.Line (-10, 0)-Step(20, 0), vbBlue
    Me.Line (0, -10)-Step(0, 20), vbBlue

    ' Draw the curve.
    y2 = F(-10)
    For X = -10 + STEP_SIZE To 10 Step STEP_SIZE
        y1 = y2
        y2 = F(X)
        Me.Line (X - STEP_SIZE, y1)-(X, y2), vbRed
    Next X
End Sub

    epsilon = -F(x) / dFdx(x)

The method then sets its next guess for x to be the current value plus epsilon (X(n+1) = X(n) + epsilon). It repeats this process until epsilon is smaller than some cutoff value.

' The function's derivative.
Private Function dFdx(ByVal X As Double) As Single
    dFdx = X * X - 4 * X
End Function

' Find the roots by using Newton's method.
Private Sub Form_MouseDown(Button As Integer, Shift As _
     Integer, X As Single,
Y As Single)
Const CUTOFF As Single = 0.0000001
Dim epsilon As Single
Dim iterations As Integer
Dim x0 As Single

    ' Clear previous results.
    DrawGraph

    x0 = X
    iterations = 0
    Do
        iterations = iterations + 1
        Me.Line (x0, -1.5)-Step(0, 3)
        epsilon = -F(x0) / dFdx(x0)
        x0 = x0 + epsilon
    Loop While Abs(epsilon) > CUTOFF
    Me.Line (x0, -3)-Step(0, 6)

    Me.Caption = x0 & " +/-" & epsilon & " in " & _
        iterations & " iterations"
End Sub

For more information on Newton's method, see Eric W. Weisstein's article Newton's Method from MathWorld--A Wolfram Web Resource.