y=f(x)
 

is it possible to solve: 0,5= x^2/(x-2)(3-0,5x)?
how can I do it?

y=f(x)
 

Use algebra, not Excel:

x = (-b +/-SQRT(b^2-4*a*c))/2a

Simplify your formula into the form

ax^2 + bx + c = 0

You'll find there are no real roots.....

HTH,
Bernie


"bbombel" wrote in message
...
is it possible to solve: 0,5= x^2/(x-2)(3-0,5x)?
how can I do it?

y=f(x)
 

Tushar Mehta has a good add-in that solves quadratics,
Once you've done what Bernie suggested (rearrange into
a quadratic) you can use the add-in to find the roots
(yes it deals with imaginary numbers...)

Check it out at
http://www.tushar-mehta.com
Under "Solve Polynomials"

http://www.tushar-mehta.com/excel/so...als/index.html

Dan E

"bbombel" wrote in message ...
is it possible to solve: 0,5= x^2/(x-2)(3-0,5x)?
how can I do it?

y=f(x)
 

Just to mention. Excel would consider the following
0.5 = x^2/(x-2)(3-0.5x)

as

0.5 = ( (x^2) / (x - 2) ) * (3 - 0.5*x)

with a real solution at 5.887850847558446

You may have meant to group the last two items as in the following
= (x^2) / ( (x - 2) * (3 - 0.5*x) )

If this is the case, you would get two imaginary solutions as Bernie
mentioned..

0.05607457622077704 - 0.580119090916421*I
0.05607457622077704 + 0.580119090916421*I
--
Dana DeLouis
Using Windows XP & Office XP
= = = = = = = = = = = = = = = = =


"bbombel" wrote in message
...
is it possible to solve: 0,5= x^2/(x-2)(3-0,5x)?
how can I do it?