Solve for x in linear regression equation

[raoulduke]

New to forum.

I have a polynomial regression equation: y=-0.0081x^2 + 0.2281x + 0.5838.

I know "y" and I have thousands of them; I want to take those values and have excel solve for "x".

Please advise! Any help would be greatly appreciated!

 

[Nebu]

Hi:

How did you predict Y which is a dependent variable without knowing x independent variable?

Thanks

 

[Chihiro]

See link for detail on how to solve for x.
https://www.mathsisfun.com/algebra/polynomials-solving.html
http://www.excel-easy.com/examples/quadratic-equation.html

Alternately you can write VBA code to simulate it.

 

[raoulduke]

Hi:

How did you predict Y which is a dependent variable without knowing x independent variable?

Thanks

Y is a variable measured every 15 minutes in the field and occasionally we measure X in the field too, but very rarely. In this case Y being depth of water and X being the amount of water.

 

[Hui]

Raoulduke

You have a quadratic equation: y=-0.0081x^2 + 0.2281x + 0.5838

Which is solved by

So your formula is rearranged to
-0.0081x^2 + 0.2281x + 0.5838-y = 0

so
a = -0.0081
b = 0.2281
c = 0.5838-y

so the two solutions are

x= [-b+sqrt(b^2-4xaxc)]/2a
&
x= [-b-sqrt(b^2-4xaxc)]/2a

You can set those up in Excel like:


see attached file:

 

[Hui]

You can then copy that down for a number of Y values