Решить уравнение численными методами

Function y(ByVal x As Double) As Double 'Функция в левой части уравнения.
    y = x ^ 5 - 2 * x ^ 4 + 2 * x ^ 3 + x ^ 2 - 7 * x - 4
End Function
 
Function dy_dx(ByVal x As Double) As Double 'Производная функции.
    dy_dx = 5 * x ^ 4 - 8 * x ^ 3 + 6 * x ^ 2 + 2 * x - 7
End Function
 
Sub Newton()
    Const baseApprox = 2# 'Начальное приближение.
    Const eps = 0.001 'Точность вычисления корня.
    Dim x0 As Double, x1 As Double
    x1 = baseApprox
    Do
        x0 = x1
        x1 = x0 - y(x0) / dy_dx(x0)
    Loop Until Abs(x0 - x1) < eps 'Повторяем итерации метода Ньютона,
                                  'пока точность не станет выше требуемой.
    MsgBox "Корень: " & x1
End Sub

Sub Coursework()
Function y(ByVal x As Double) As Double
       y = x ^ 5 - 2 * x ^ 4 + 2 * x ^ 3 + x ^ 2 - 7 * x - 4
End Function
Function dy_dx(ByVal x As Double) As Double
   dy_dx = 5 * x ^ 4 - 8 * x ^ 3 + 6 * x ^ 2 + 2 * x - 7
End Function
Sub Newton()
    Const baseApprox = 2#
    Const eps = 0.001
    Dim x0 As Double, x1 As Double
    x1 = baseApprox
    Do
x0 = x1
        x1 = x0 - y(x0) / dy_dx(x0)
    Loop Until Abs(x0 - x1) < eps
MsgBox "Êîðåíü: " & x1
End Sub
Sub Newton2()
Const baseApprox = 0#
    Const eps = 0.001
    Dim x0 As Double, x2 As Double
    x2 = baseApprox
    Do
x0 = x2
        x2 = x0 - y(x0) / dy_dx(x0)
    Loop Until Abs(x0 - x2) < eps
MsgBox "Êîðåíü: " & x2
End Sub
Sub Newton3()
Const baseApprox = -1#
    Const eps = 0.001
    Dim x0 As Double, x3 As Double
    x3 = baseApprox
    Do
x0 = x3
        x3 = x0 - y(x0) / dy_dx(x0)
    Loop Until Abs(x0 - x3) < eps
MsgBox "Êîðåíü: " & x3
End Sub
Sub Newton4()
Const baseApprox = 3#
    Const eps = 0.001
    Dim x0 As Double, x4 As Double
    x4 = baseApprox
    Do
x0 = x4
        x4 = x0 - y(x0) / dy_dx(x0)
    Loop Until Abs(x0 - x4) < eps
MsgBox "Êîðåíü: " & x4
End Sub