(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
What you wrote is, apparently, $\displaystyle \displaystyle{z+h+\left[\frac{Q}{b}\cdot\frac{(z+h)^2}{2g}\right]}=z\Longrightarrow \displaystyle{h+\left[\frac{Q}{b}\frac{(z+h)^2}{2g}\right]}=0\Longrightarrow \displaystyle{(z+h)^2=-\frac{2bgh}{Q}}$ ...
Yet the above is so simple that perhaps you meant another first expression...
Tonio
$\displaystyle \displaystyle z+h+\left(\frac{Q}{b(z+h)}\right)^2\times\frac{1}{ 2g}=z$
Subtract $\displaystyle z$ from both sides
$\displaystyle \displaystyle h+\left(\frac{Q}{b(z+h)}\right)^2\times\frac{1}{2g }=0$
Take $\displaystyle h$ from both sides
$\displaystyle \displaystyle \left(\frac{Q}{b(z+h)}\right)^2\times\frac{1}{2g}=-h$
Now lets simplify this a bit
$\displaystyle \displaystyle \frac{Q^2}{b^2(z+h)^2}\times\frac{1}{2g}=-h$
$\displaystyle \displaystyle \frac{Q^2}{2gb^2(z+h)^2}=-h$
Now your turn, multiply both sides through by $\displaystyle \displaystyle (z+h)^2$