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Sharp el 510r

How do i plug in quadratic formula

Posted by Anonymous on

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Posted on Jan 02, 2017

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

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

Hello,
The calculator does not have a solve program, but you can still use it to solve the quadratic equation with a little effort. If you know the therory skip toward the end.

Let aX^2+bX+C=0
1. First calculate the discriminant usually called Delta and given by
Delta =b^2 -4*a*c store the value in the variable D
If Delta is positive you have two roots X1 and X2 given by
X1=(-b+square root of Delta)/(2*a)
X2=(-b- square root of Delta)/(2*a)

If Delta is equal to 0, X1=X2=-b/(2*a)
If Delta i negative, no real solutions exist

You only solve if Delta is positive or equal to 0.

Putting in the values of a, b, and c
value of a (put you number) [SHIFT][STO] A
value of b (put your value) [SHIFT][STO] B
value of c (put you value) [SHIF][STO] C

[ALPHA] B [X^2] -4[x][ALPHA] A [x][ALPHA] C [=]
Value of delta is displayed. If it is positive, you store its square root in D

[Square root] [ANS] [SHIFT][STO] D ;

Calculate 1/(2*a) and store in variable F
1./(2[x] ALPHA A) [SHIFT][STO] F

To obtain X1
[ALPHA] F [x] ( (-) ALPHA B + ALPHA D ) [=]
To obtain X2
[ALPHA] F [x] ( (-) ALPHA B - ALPHA D ) [=]

Be careful: the (-) is the change sign not the regular minus sign.

Hope it helps.

Posted on Sep 24, 2009

Whay would you want to go through the hassle of writing a program to solve to solve a quadratic equation`The calculator has two buit-in command to do just that. It has Solve and cSolve (for complex solutions).

Open a calculator sheet, and on the command line, type in solve(ax^2+bx+c=0, x), press ENTER and wait for the solutions

Posted on May 11, 2011

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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.

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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.

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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.

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