### Question

• I searched Help and came across some rhetoric about Quadratic Equations (MS Excel 2007 in Office 2007) but no description as to how to solve the equation.

I installed the Solver Add-In and still can't get it going. There is a box which opens but there are no indications as to how to use it and nothing looks relative to quadratic equations.

I did find online from Google a page from a book which shows how to build a formula.

But I'd really like to see the integrated solver do what is paid to do - solve the equation.

Does anyone on here know how to do it?

TIA

Monday, May 31, 2010 1:41 AM

• I came across this:

 QUADRATIC EQUATION SOLVER a b c disc 4.9 17 -42 1112.2 Root1 1.668333 Root2 -5.13772

Where cell E4 is =b*b-4*a*c_

cell B7 is =(-b+SQRT(disc))/(2*a)

and cell D7 is =(-b-SQRT(disc))/(2*a)

It returns both roots.

It's not perfect by any means. There are problems with negative numbers and with small numbers.

With all the formulas and equations and calculations in Excel it seems quite unreasonable they don't have this one nailed down.

Thanks to all who responded to my request.

• Marked as answer by Tuesday, June 01, 2010 3:03 AM
Tuesday, June 01, 2010 3:03 AM

### All replies

• See http://www.solver.com/suppstdsolver.htm for detailed help on how to use the Solver Add-In.
Simon Jones http://pcpro.co.uk
Monday, May 31, 2010 9:56 PM
• Thanks for the reply Simon. I had headed towards Solver because when I opened Excel Help (after I found where they hid it) and typed in Quadratic Equation it mentioned Solver. So I plugged in the Add-On (after figuring out where they hid that) and can't seem to make anything of it (not a surprise since everything else that was simple in 2003 is impossible in 2007) and so I guess I'll just throw this junk software away and go with Open Office.

Monday, May 31, 2010 10:29 PM
• Provided that the quadratic equation has at least one real root, you can use the solver as follows:

then, in column A, set A1 to the value a, A2 to b, A3 to c and put a starting guess for the root in A4 (or leave it blank which will be read as 0).
Then in B1, type the formula =A1*A4*A4+A2*A4+A3.

Go to tools, solver. In set target cell box type B1. Set equal to value 0. In the by changing cells box type A4, click solve. The value in A4 is the root it found. You might also want to click on options and set the tolerance to a suitably small value before clicking solve. You can see how close to 0 the quadratic ezpression in B1 has got as it displays the result.

It only finds one root.

Or, of course, you can use the equation for the roots of a quadratic directly (in any elementary maths book).
"Simon Jones [PC Pro]" <=?utf-8?B?U2ltb24gSm9uZXMgW1BDIFByb10=?=> wrote in message news:3f3da9c2-b8f2-4328-89f5-3a0e7106dd49...
See http://www.solver.com/suppstdsolver.htm for detailed help on how to use the Solver Add-In.
Simon Jones http://pcpro.co.uk
Tuesday, June 01, 2010 1:03 AM
Actually, if it has two real roots  you can, by setting a suitable constraint find the second one also (only one at a time, though).
"JRosenfeld" <=?utf-8?B?SlJvc2VuZmVsZA==?=> wrote in message news:abca1757-bbd9-4ece-bd79-5ce7bdba9c5e...
Provided that the quadratic equation has at least one real root, you can use the solver as follows:

then, in column A, set A1 to the value a, A2 to b, A3 to c and put a starting guess for the root in A4 (or leave it blank which will be read as 0).
Then in B1, type the formula =A1*A4*A4+A2*A4+A3.

Go to tools, solver. In set target cell box type B1. Set equal to value 0. In the by changing cells box type A4, click solve. The value in A4 is the root it found. You might also want to click on options and set the tolerance to a suitably small value before clicking solve. You can see how close to 0 the quadratic ezpression in B1 has got as it displays the result.

It only finds one root.

Or, of course, you can use the equation for the roots of a quadratic directly (in any elementary maths book).
"Simon Jones [PC Pro]" <=?utf-8?B?U2ltb24gSm9uZXMgW1BDIFByb10=?=> wrote in message news:3f3da9c2-b8f2-4328-89f5-3a0e7106dd49...
See http://www.solver.com/suppstdsolver.htm for detailed help on how to use the Solver Add-In.
Simon Jones http://pcpro.co.uk
Tuesday, June 01, 2010 1:22 AM
• I came across this:

 QUADRATIC EQUATION SOLVER a b c disc 4.9 17 -42 1112.2 Root1 1.668333 Root2 -5.13772

Where cell E4 is =b*b-4*a*c_

cell B7 is =(-b+SQRT(disc))/(2*a)

and cell D7 is =(-b-SQRT(disc))/(2*a)

It returns both roots.

It's not perfect by any means. There are problems with negative numbers and with small numbers.

With all the formulas and equations and calculations in Excel it seems quite unreasonable they don't have this one nailed down.

Thanks to all who responded to my request.

• Marked as answer by Tuesday, June 01, 2010 3:03 AM
Tuesday, June 01, 2010 3:03 AM
• Yes, well that is just using the algebraic expression for the roots of a quadratic equation, which I had suggested:
roots = (-b+/-sqrt(b*b -4ac)/2a
"hmmmm....." <=?utf-8?B?aG1tbW0uLi4uLg==?=> wrote in message news:bcf96392-2342-4bcd-af29-d04f59877dd4...

I came across this:

 QUADRATIC EQUATION SOLVER a b c disc 4.9 17 -42 1112.2 Root1 1.668333 Root2 -5.13772

Where cell E4 is =b*b-4*a*c_

cell B7 is =(-b+SQRT(disc))/(2*a)

and cell D7 is =(-b-SQRT(disc))/(2*a)

It returns both roots.

It's not perfect by any means. There are problems with negative numbers and with small numbers.

With all the formulas and equations and calculations in Excel it seems quite unreasonable they don't have this one nailed down.

Thanks to all who responded to my request.

Wednesday, June 02, 2010 12:51 AM