# rearranging formula

• November 22nd 2010, 09:48 AM
Haris
rearranging formula
(z+h)+[(Q/b*(z+h))^2/(2*g)]=z

Could someone please solve for z.

We know all the variables apart from z.

Cheers :)
• November 22nd 2010, 04:43 PM
topspin1617
Well, this is the best I can make out from how you have the original problem written... learning to write this stuff in TEX on here can be helpful :)

$z+h+\frac{(\frac{Q}{b}(z+h))^2}{2g}=z$.

Is this the correct problem?

Btw, I don't think this belongs in this section... this really feels like more of a basic algebra question than abstract algebra or whatever.

But I'll help anyway since it's here.

What work have you been able to do yourself so far?
• November 22nd 2010, 06:22 PM
mr fantastic
Quote:

Originally Posted by Haris
(z+h)+[(Q/b*(z+h))^2/(2*g)]=z

Could someone please solve for z.

We know all the variables apart from z.

Cheers :)

If you introduce some new notation such as x = z + h and (Q/b)^2/2g = a it's much easier to see:

$x + ax^2 = x - h \Rightarrow ax^2 = -h$. If a < 0 then there is no real solution for x and hence z.