Have looked in the user guide – cannot find information on use of the quadratic formula.
[Math] Casio fx-85GT PLUS calculator. How to access & then use the quadratic equation formula
algebra-precalculuscalculator
Related Solutions
I would like to prove the Quadratic Formula in a cleaner way. Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula.
Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.)
We want to solve the equation $$ax^2+bx+c=0,$$ where $a \ne 0$. The usual argument starts by dividing by $a$. That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset.
Instead, multiply both sides by $4a$. We obtain the equivalent equation
$$4a^2x^2 +4abx+4ac=0.\tag{1}$$
Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely,
$$4a^2x^2+4abx=(2ax+b)^2-b^2.$$
So our equation can be rewritten as
$$(2ax+b)^2 -b^2+4ac=0 \tag{2}$$
or equivalently
$$(2ax+b)^2=b^2-4ac. \tag{3}$$
Now it's all over. We find that
$$2ax+b=\pm\sqrt{b^2-4ac} \tag{4}$$
and therefore
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \tag{5}$$
No fractions until the very end!
Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form.
Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots. Ideally, each type of proof should be presented, since each connects to an important family of ideas. And a twice proved theorem is twice as true.
The symbol $\pm$ ("$+$ or $-$") means that either $+$ or $-$ will produce a solution - so use each, getting a different number each time, and you will get two answers.
For example: say I've solved an equation and gotten $x = 2 \pm 7$. Then $2 - 7 = -5$ and $2 + 7 = 9$ are both solutions to my equation. What that means is that if I've only been asked for "a" solution, I can give either $-5$ or $9$ as my answer - both are correct. But if I've been asked for "the" solution or "all" solutions, then the correct answer is "$x = -5$ or $x = 9$".
Your particular example looks like $\frac{6 \pm \sqrt{36 - 4 \cdot 4.0 \cdot \pm 7.04}}{14.2}$. If that's correct, then there are two $\pm$'s to deal with, and these can be evaluated independently. For example, if I were looking at $1 \pm 2 \pm 3$, then there would be four possible solutions: $1 + 2 + 3$, $1 + 2 - 3$, $1 - 2 + 3$, and $1 - 2 - 3$.
Word of warning: I said "if that's correct" because it doesn't look like it is. You referenced the quadratic formula, but if that's the only place $\pm$'s are coming from, there shouldn't be two of them. The quadratic formula states that the solutions to the quadratic $ax^2 + bx + c = 0$ are $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$; notice that there's only one $\pm$ in that.
Best Answer
Most likely, the Casio fx-85GT PLUS (as with virtually all scientific calculators) doesn't do anything special for the quadratic formula. If you know the quadratic formula, you can substitute the values of the coefficients of your quadratic and then enter the resulting expressions into your calculator.