# Math Help - Reverse calculation foruma for simple lookup

1. ## Reverse calculation foruma for simple lookup

I am building an app for a client and not exactly a math wiz. I've been able to get pretty far using formulas they've given me, but now I need to reverse find a value based on some constants I already have.

first find for x:
15 = (x/0.6)squared * 40

then using x… find y:
x = (y*20*15)/5940

2. ## Re: Reverse calculation foruma for simple lookup

Originally Posted by bdub01
first find for x:
15 = (x/0.6)squared * 40
then using x… find y: x = (y*20*15)/5940
This a a bit hard to read.
But if you know that $15 = 40\left( {\frac{x}{{0.6}}} \right)^2$ then there are two solutions:
$x = \pm \left( {0.6} \right)\sqrt {\frac{{15}}{{40}}}$

I suspect that you need only the positive solution here.

3. ## Re: Reverse calculation foruma for simple lookup

Originally Posted by bdub01
then using x… find y:
x = (y*20*15)/5940
x = 300y / 5940
x = 10y / 198
10y = 198x
y = 19.8x