Question about Sharp EL-1801C Calculator

# I have a sharp EL-1801C and having trouble with basic addition and subtraction. Was wondering if it could be a setting problem? For example: 100+50=100 or the answer always is the first number of the equation. Please help ASAP! Getting very frustrated and need to use it as soon as possible. Thank You, Rita Gosselin

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## 5 Suggested Answers

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Posted on Jan 02, 2017

To change the date, enter the date in MM.DD.YY format, pressing the decimal key between the month and day (will show as a decimal on the display) and the day and year (will show as a dash on the display). Then, press the Time/Date key (the one that looks like a clock and a calendar). The new date will show on the display. Note that if the month is less than 10, you do need to press the 0 (zero) key for the first digit, even though it will not show on the display. To change the time, enter the time in 24-hour HH.MM format, pressing the decimal key between the hour and minute. Then, press the Time/Date key. The new time will show on the display. Note that if the hour is less than 10, you do need to press the 0 (zero) key for the first digit, even though it will not show on the display. Hope this helps! :)

Posted on Dec 04, 2008

Hello,
This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.
Let the equation be ax^2 +bx+c=0
You compute the discriminant Callit disc.
disc= (b^2-4*a*c)
1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.
Then x1=(-b+square root of disc)/(2*a)
and x2= (-b-square root of disc)/(2*a)

2. If disc=0 , square root of disc =0 and x1=x2=-b/(2*a)
3. If disc is negative, there are no real solutions.

To find the solutions you replace a, b, and c by their numerical values, to calculate disc, its square root, and calculate X1 and X2.

Hope it helps.

Posted on Sep 17, 2009

Hello,
This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.
Let the equation be ax^2 +bx+c=0
You compute the discriminant Call it disc.
disc= (b^2-4*a*c)
1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.
Then x1=(-b+square root of disc)/(2*a)
and x2= (-b-square root of disc)/(2*a)

2. If disc=0 , square root of disc =0 and x1=x2=-b/(2*a)
3. If disc is negative, there are no real solutions.

To find the solutions you replace a, b, and c by their numerical values, to calculate disc, its square root, and calculate X1 and X2.

Hope it helps.

Posted on Sep 17, 2009

Here is the manual on link below on page 5 tells you how to replace.

http://dl.owneriq.net/e/ea19126b-3714-4b3d-bce5-712458b60c6d.pdf

Good luck and please rate 4 thumbs for the free answer

Posted on Mar 14, 2010

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Here is the manual on link below on page 5 tells you how to replace.

http://dl.owneriq.net/e/ea19126b-3714-4b3d-bce5-712458b60c6d.pdf

Good luck and please rate 4 thumbs for the free answer

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