Quadratic Formula help in vba

[vop2311]

Hi Experts,

I need some help to fix my Quadratic Equation in VBA. It looks fine except sqrt part.

Here is the normal equation below X =

-b +- √b^2-4(a)(c)
---------------------
2(a)


Here is my b value
Moderator's note: Please take the time to review our rules. There aren't many, and they are all important. Rule #3 requires code tags. I have added them for you this time because you are a new member. --6StringJazzer

b = (Range("C13") * Range("C14") / Range("C14") + vt * (xf - 1)

Based on B my equation looks like this

X = - Range("C13") * Range("C14") / Range("C14") + vt * (xf - 1) - Sqr(Range("C13") * Range("C14") / Range("C14") + vt) * (xf - 1) ^ 2 - 4 * ksp / 2

It would be great if someone can give an idea on sqrt part. Beside that all seems to be working.

Thanks,
Vik

[6StringJazzer]

You are on the right track. The main problem is that your parentheses are in the wrong place. But you have bigger problems than that so I'm going to make you start from scratch.

I highly recommend that you declare variables a, b, and c. Then your formula is much more readable, and easier to set up and debug. It will look exactly like the quadratic formula, in code form.

The quadratic formula gives two roots, but you have only calculated one of them. I am providing code to calculate both.

Your assignment to b has a syntax error. Remove the initial left parenthesis.

From context, it looks to me like c = ksp and a = 1. Never imply variables. If it's 1, then show a 1.

Here is the best practice for coding this:

   Dim a As Double
   Dim b As Double
   Dim c As Double
   
   Dim vt As Double  ' no idea what this is
   Dim xf As Double  ' no idea what this is
   Dim ksp As Double ' no idea what this is
   
   Dim x1 As Double ' first root
   Dim x2 As Double ' second root
   
   a = 1
   c = ksp
   b = Range("C13") * Range("C14") / Range("C14") + vt * (xf - 1)

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

[protonLeah]

double post

[6StringJazzer]

Welcome to the Forum, unfortunately:

This is a duplicate post and as such does not comply with Rule 5 of our forum rules. This thread will now be closed, you may continue in your other thread.

Thread Closed.