# Math Problem • ### Question

• I have got this to give the right answer in excel but not in vb.

Did I do this wrong?

Excel I get 106.42

VB I get NaN

Try

Dim a1 As Double = TextBox1.Text

Dim b1 As Double = TextBox2.Text

Dim c1 As Double = TextBox3.Text

'ROUNDUP(SUM(SUM(2*B9)+1.57*SUM(B6+B3))+SUMSQ(B3-B6)/SUM(4*B9),2)

Label9.Text = ((2 * c1) + 1.57 * (b1 + a1)) + Math.Sqrt(a1 - b1) / (4 * c1)

Catch ex As Exception

End Try

Thanks

David

Thursday, March 20, 2008 8:41 PM

• If you're wanting to get the square root in excel, replace SUMSQ with SQRT.

If you're wanting to get the square in VB, use the following code:

Code Snippet

Label9.Text = ((2 * c1) + 1.57 * (b1 + a1)) + (a1 - b1)^2 / (4 * c1)

Thursday, March 20, 2008 9:47 PM

### All replies

• Provide the values you plugged in.

Thursday, March 20, 2008 9:23 PM
• Scratch that, I see your problem.

SUMSQ returns the sum of the squares in the argument list (in this case, the square of  B3-B6).

In your VB code, you're finding the square root, which is not a number (well, technically an imaginary number) if b1 is greater than a1.

Thursday, March 20, 2008 9:31 PM
• Dim a1 As Double = TextBox1.Text    '=7

Dim b1 As Double = TextBox2.Text    '=17

Dim c1 As Double = TextBox3.Text    '=34

'ROUNDUP(SUM(SUM(2*B9)+1.57*SUM(B6+B3))+SUMSQ(B3-B6)/SUM(4*B9),2)

Label9.Text = ((2 * c1) + 1.57 * (b1 + a1)) + Math.Sqrt(a1 - b1) / (4 * c1)

Catch ex As Exception

End Try

I am confused, how do you get the square root?

Davids Learning

Thursday, March 20, 2008 9:35 PM
• If you're wanting to get the square root in excel, replace SUMSQ with SQRT.

If you're wanting to get the square in VB, use the following code:

Code Snippet

Label9.Text = ((2 * c1) + 1.57 * (b1 + a1)) + (a1 - b1)^2 / (4 * c1)

Thursday, March 20, 2008 9:47 PM
• Thanks Chris

The ^ did the trick.

Davids Learning

Thursday, March 20, 2008 10:09 PM