# Thread: Rearranging a Difficult Equation

1. ## Rearranging a Difficult Equation

Hi all,

I've been trying (unsuccessfully!) to rearrange the following equation so that H is the subject:

mgH = [k(H - L)^2] / 2

(Where ^2 means squared.) I also have some test data which shows that if m = 50, g = 9.8, k = 40 and L = 20, H should equal 57.5, or thereabouts.

I'm going to need the equation in a couple of hours and am stressing out, so any help you can offer would be greatly appreciated.

2. Originally Posted by mothonthewall86
Hi all,

I've been trying (unsuccessfully!) to rearrange the following equation so that H is the subject:

mgH = [k(H - L)^2] / 2

(Where ^2 means squared.) I also have some test data which shows that if m = 50, g = 9.8, k = 40 and L = 20, H should equal 57.5, or thereabouts.

I'm going to need the equation in a couple of hours and am stressing out, so any help you can offer would be greatly appreciated.
$\displaystyle mgH = \frac{k}{2}(H^2 - 2HL + L^2)$

$\displaystyle mgH = \frac{k}{2}H^2 - kHL + \frac{k}{2}L^2$

$\displaystyle 0 = \frac{k}{2}H^2 - kHL - mgH + \frac{k}{2}L^2$

$\displaystyle 0 = \frac{k}{2}H^2 - (kL + mg)H + \frac{k}{2}L^2$

use the quadratic formula ... $\displaystyle H = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ , where ...

$\displaystyle a = \frac{k}{2}$ , $\displaystyle b = -(kL + mg)$ , $\displaystyle c = \frac{k}{2}L^2$

3. $\displaystyle mgH = \frac{k(H - L)^2}{2}$

Multiply both sides by 2

$\displaystyle 2mgH = k(H - L)^2$

Multiply out the brackets

$\displaystyle 2mgH = k(H^2 - 2LH + L^2)$ = $\displaystyle 2mgH = kH^2 - 2LkH + kL^2$.

You now have a quadratic, rearrange and solve for H.

4. I must be dense for not seeing that. Internet strangers, you are my knights in shining armor!

Thanks so much for your help.