# Need help with quadratic equation **URL:** \ **Category:** Factual Questions\ **Created:** [April 20, 2006, 12:24am UTC](https://boards.straightdope.com/t/need-help-with-quadratic-equation/353394 "2006-04-20T00:24:06Z")\ **Posts on this page:** 9\ **Page:** 1
**Author:** ![Vlad\_Igor](https://avatars.discourse-cdn.com/v4/letter/v/7feea3/32.png) [@Vlad\_Igor](https://boards.straightdope.com/u/Vlad_Igor)\ **Post date:** [April 20, 2006, 12:24am UTC](https://boards.straightdope.com/t/need-help-with-quadratic-equation/353394/1 "2006-04-20T00:24:06Z")
I’ve been busy building an enzymatic assay to measure a protein. The data points of the standards used to create a calibration curve, when graphed, fit perfectly on a second order polynomial curve (i.e. aX^2+bX+c). I remember the formula for solving for X when Y=0, however, we’re in the real world, and I need to find X when, say, 1.225=aX^2+bX+c. From Excel, I know a, b, and c, and from my experimental data, I know Y. What I need to know, and can’t find, is how to solve for X when Y has an explicit value not equal to 0. Any help from those more mathematically inclined than I would be greatly appreciated. Vlad/Igor ---
**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\ **Post date:** [April 20, 2006, 12:26am UTC](https://boards.straightdope.com/t/need-help-with-quadratic-equation/353394/2 "2006-04-20T00:26:16Z")
> [@Vlad/Igor](#): > > I’ve been busy building an enzymatic assay to measure a protein. The data points of the standards used to create a calibration curve, when graphed, fit perfectly on a second order polynomial curve (i.e. aX^2+bX+c). I remember the formula for solving for X when Y=0, however, we’re in the real world, and I need to find X when, say, 1.225=aX^2+bX+c. From Excel, I know a, b, and c, and from my experimental data, I know Y. What I need to know, and can’t find, is how to solve for X when Y has an explicit value not equal to 0. Any help from those more mathematically inclined than I would be greatly appreciated. > > Vlad/Igor Set the equation to aX^2+bX+(c-Y)=0 and use the QE ---
**Author:** ![Vlad\_Igor](https://avatars.discourse-cdn.com/v4/letter/v/7feea3/32.png) [@Vlad\_Igor](https://boards.straightdope.com/u/Vlad_Igor)\ **Post date:** [April 20, 2006, 3:11pm UTC](https://boards.straightdope.com/t/need-help-with-quadratic-equation/353394/3 "2006-04-20T15:11:48Z")
I tried that and got the following, but I don’t know if it correct: X = -b+/- sqrt(4a(Y-c)+b^2)/2a Vlad/Igor ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 20, 2006, 3:19pm UTC](https://boards.straightdope.com/t/need-help-with-quadratic-equation/353394/4 "2006-04-20T15:19:43Z")
That’s right. You can always substitute the values back in the original equation to check. ---
**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\ **Post date:** [April 20, 2006, 5:28pm UTC](https://boards.straightdope.com/t/need-help-with-quadratic-equation/353394/5 "2006-04-20T17:28:03Z")
Your answer is right, assuming that you implicitly have parentheses as follows: X = (-b+/- sqrt(4a(Y-c)+b^2))/(2a) That is, the 2a needs to divide the entire sum -b+/- sqrt(4a(Y-c)+b^2), not just the square root. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [April 20, 2006, 9:54pm UTC](https://boards.straightdope.com/t/need-help-with-quadratic-equation/353394/6 "2006-04-20T21:54:46Z")
Shouldn’t it be x=(-b+/-sqrt(b[sup]2[/sup]\*\*-\*\*4_a_c))/(2a) where c=c-Y? ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 21, 2006, 12:35am UTC](https://boards.straightdope.com/t/need-help-with-quadratic-equation/353394/7 "2006-04-21T00:35:15Z")
> [@treis](#): > > Shouldn’t it be x=(-b+/-sqrt(b[sup]2[/sup]\*\*-\*\*4_a_c))/(2a) where c=c-Y? c = c - Y implies Y = 0. ---
**Author:** ![Larry\_Borgia](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/larry_borgia/32/156_2.png) [@Larry\_Borgia](https://boards.straightdope.com/u/Larry_Borgia)\ **Post date:** [April 21, 2006, 12:43am UTC](https://boards.straightdope.com/t/need-help-with-quadratic-equation/353394/8 "2006-04-21T00:43:01Z")
> [@ultrafilter](#): > > c = c - Y implies Y = 0. I think he meant a new c, c[sub]2[/sub], where c[sub]2[/sub] = c[sub]1[/sub] - Y ---
**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\ **Post date:** [April 21, 2006, 1:04am UTC](https://boards.straightdope.com/t/need-help-with-quadratic-equation/353394/9 "2006-04-21T01:04:56Z")
> [@treis](#): > > Shouldn’t it be x=(-b+/-sqrt(b[sup]2[/sup]\*\*-\*\*4_a_c))/(2a) where c=c-Y? Yes, but that’s equivalent because replacing c with Y-c instead of c-Y cancels the minus sign.