Question

编写函数求一元二次方程ax2 +bx+c=0的解，分别通过文本框输入系数a，b，c，输出解x1，x2。

Answer 1

以下是核心算法：其中Text1，Text2，Text3是三个文本输入框，接受a，b，c三个系数。x1，x2为根
Dim a As Single, b As Single, c As Single, d As Single, ds As Single
Dim x1 As String, x2 As String

a = Val(Text1.Text)
b = Val(Text2.Text)
c = Val(Text3.Text)
d = b * b - 4 * a * c
ds = Sqr(Abs(d))
If d > 0 Then
x1 = Str((-b + ds) / 2 * a)
x2 = Str((-b - ds) / 2 * a)
ElseIf d = 0 Then
x1 = Str(-b / 2 * a): x2 = x1
Else
b = Round(-b / (2 * a), 7): c = Round(ds / (2 * a), 7)
x1 = Str(b) + "+" + Str(c) + "i"
x2 = Str(b) + "-" + Str(c) + "i"
End If

——————————————————————转贴

Answer 2

x1=(-b+sqrt(b*b-4*a*c))/(2*a)
x2=(-b-sqrt(b*b-4*a*c))/(2*a)