Solving a Quadratic Equation

by Allen Wyatt
(last updated October 12, 2019)

2

Clyde wonders if there is a function in Excel that solves quadratic equations. The short answer is that there is no such worksheet function built into the program. You can, however, use a couple of formulas to solve a quadratic equation, or you can develop your own user-defined function.

This is the general form of a quadratic equation:

ax^2 + bx + c = 0

In the equation, values a, b, and c are constants and you need to solve for x. There are always two possible solutions for x, which means that there are two quadratic formulas that derive from the quadratic equation:

(-b+SQRT(b^2-4ac))/2a
(-b-SQRT(b^2-4ac))/2a

This is easy enough to place into a worksheet. Assuming that your values for a, b, and c are in cells A1, B1, and C1, you could use the following formulas:

=(-B1+SQRT(B1^2-4*A1*C1))/(2*A1)
=(-B1-SQRT(B1^2-4*A1*C1))/(2*A1)

You can, at this point, change the values in A1, B1, and C1 and end up with the answers you need. These formulas will work for any value where (B1^2)>(4*A1*C1). If this is not true, there are still answers, but you start getting into imaginary numbers. At this point things can get sticky in a hurry, and mere mortals run into the real possibility of their heads exploding. If you want to adjust your formulas to deal with imaginary numbers, you might find this short discussion helpful:

https://www.excelforum.com/excel-formulas-and-functions/1116652-how-to-view-complex-number-from-quadratic-equation.html

I should note that I've also seen people use Excel's Goal Seek tool to solve a quadratic equation, but for my money, the formulas described above work just fine. My feeling is that if you can use a formula to derive a solution, it seems to me that the whole need for doing the iterative work at which the Goal Seek tool is so adept is removed. If you prefer to work with the Goal Seek tool, however, you may find this web page helpful:

https://www.wikihow.com/Solve-a-Quadratic-Equation-Using-the-Goal-Seek-Feature-on-Microsoft-Excel

I mentioned earlier that you could build your own user-defined function to derive your solutions to a quadratic equation. Here is one that you could easily implement:

Function Quadratic(a As Double, b As Double, _
  c As Double, r As Integer) As Variant

    Quadratic = CVErr(xlErrValue)
    If r = 1 Then
        Quadratic = CVErr(xlErrNA)
        If b ^ 2 > (4 * a * c) Then
            Quadratic = (-b + Sqr(b ^ 2 - (4 * a * c))) / (2 * a)
        End If
    End If
    If r = 2 Then
        Quadratic = CVErr(xlErrNA)
        If b ^ 2 > (4 * a * c) Then
            Quadratic = (-b - Sqr(b ^ 2 - (4 * a * c))) / (2 * a)
        End If
    End If
End Function

The function requires four parameters: the values for A, B, and C, plus an indicator of whether you want the first (positive) answer or the second (negative) answer. Given three values in cells A1, B1, and C1, here's how you would get the second (negative) result in your worksheet:

=Quadratic(A1, B1, C1, 2)

If your indicator (the fourth parameter) is not 1 or 2, then the function returns a #VALUE! error. If the constants you provide for the first three parameters result in an imaginary answer, then the function returns a #NUM! error.

Note:

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ExcelTips is your source for cost-effective Microsoft Excel training. This tip (13686) applies to Microsoft Excel 2007, 2010, 2013, 2016, 2019, and Excel in Office 365.

Author Bio

Allen Wyatt

With more than 50 non-fiction books and numerous magazine articles to his credit, Allen Wyatt is an internationally recognized author. He is president of Sharon Parq Associates, a computer and publishing services company. ...

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What is 1 + 3?

2019-10-14 11:31:09

Roy

Also, there is an "inverse" (probably there is a better word, I got that one with Goal Seek...) form of the equation that used to be important given the calculational tools of the day (pen and paper, sliderules).

Its use was suggested when "a" in the equation was very small compared to the other values. Given Excel's difficulties with digits past 14 or so (binary math), one could probably find that one better to use for what is effectively the same reason. A simple IF() test could let your formula select between the usual version and that one. Again, the point would be to work within the accuracy of the calculational tool, not that either is right or wrong (one just uses algebra to shift things about, rearrange them, not use a whole different formula, so essentially it is the same formula, just rearranged).


2019-10-14 11:23:49

Roy

Yes, using the QF is ALWAYS the way to go.

Goal Seek's only role in life is for things that don't have precise formulas or require math beyond your skill set and you can't find an exact solution which you can adapt, usually for the same reason, skill set-wise.

That's because it simply isn't precise and it's certainly not a two second throwaway thing like typing =54+82 instead of firing up a calculator or doing it in your head is. So if a precise formula exists which requires no real adaptation to your needs...

By the way, if that seems harsh on Goal Seek, part of it is that I have never seen a real question that could use Goal Seek that wasn't a troll where the questioner will preach about it if no answerer does. Only professional Excel people seem to treat it as something to solve real things, rather than something to solve anything and everything. The other part is that using Goal Seek here is just as... sad... as using handwritten graphing to "solve" the equation. Only matches the right answer by accident. The fact that one worked to get that accident doesn't make it "correct." Since the QF is simple, needs no real adaptation, and one can simply round at whatever digit one pleases, why use a hit-or-miss method that is more work? So... harsh on good ol' Goal Seek. Use it where it is STRONG, not where it is lucky at best and more trouble anyway!


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