How would you use the quadratic formula to solve x^2+5x=-2. Using compete sentences.

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How would you use the quadratic formula to solve x^2+5x=-2. Using compete sentences.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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maybe we can work the prob using math, and you do the sentences part alone?
Okay

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when the equation is \(\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\). you use the following formula: \(\normalsize\color{ slate }{\Huge{\bbox[5pt, lightcyan ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{-\color{magenta}{b} \pm\sqrt{ \color{magenta}{b} ^2-4 \color{blue}{a} \color{red}{c}}}{2 \color{blue}{a}} }~ }}}\) (to transfer your equation to the needed \(\large \color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\) form, you have to add 2 to both sides.)
So it will be x^2+5+2=0
what do you get after adding 2 to both sides?
yes
\(\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\) ` ... comparing ... ` \(\LARGE\color{black}{ \color{blue}{1} x^2+ \color{magenta}{5}x+ \color{red}{2}=0 }\)
Now, please plug in the a b c for me....
What do you mean by plug in?
I have no clue how to do this. I am just learning it
she meant let a = 1, b = 5, and c = 2 we will have this equation \[x^2+5x+2=0\]
Okay and then what do I do from there?
Well, to solve this faster we need the discriminant \[b^2-4ac\] so let b = 5, a =1, and c =2
who is she?
umm I thought that was you?
well, if you are thinking of me as of a good looking girl, rather than a bad looking girl, then you in a sense did tackle my identity.
anyway plug in b =5, c =2, and a = 1 into the discriminant formula \[b^2-4ac\] @J_slate23
as in replace the b with 5, replace the c with 2 and replace the a with 1
So 5^2=4(1)(2)?
yes but replace the = with -
\[5^2-4(1)(2)\]
So 5^2-4(1)(2)?
mhm so now what's 5^2 and what's 4(1)(2)
\(\normalsize\color{ slate }{\Huge{\bbox[5pt, lightyellow ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{-\color{magenta}{5} \pm\sqrt{ \color{magenta}{5} ^2-4 \color{blue}{(1)} \color{red}{(2)}}}{2 \color{blue}{(1)}} }~ }}}\) and your equation is \(\LARGE\color{black}{ \color{blue}{1} x^2+ \color{magenta}{5}x+ \color{red}{2}=0 }\).
25 and 8
yes so now what's 25-8
17
yes! so for the bottom denominator what is 2(1) ?
2
yes
so now you have \[\frac{5 +\sqrt{17}}{2}, \frac{5-\sqrt{17}}{2}\] I split the signs up because the latex for the quadratic formula is big and nasty
OH MAN I forgot the -!
\[\frac{-5 +\sqrt{17}}{2}, \frac{-5-\sqrt{17}}{2}\]
Is that the answer? If so that was really easy
yeah.. I just split the sign in the middle up. the latex for the entire quadratic equation with + and - together was a pain
and 17 is not a perfect square... so we can't simplify .. not to mention 17 is a prime
So the final answer is -5+-sqrt17/2?
yeah
Thank you!
That is easier then I expected
Can you check my math for another one? @UskiDoll
sure
Solve x^2+4x-12=0 by completing the square. X^2+4x-12=0 Add 12 X^2+4x=12 (X+2)^2-4=12 (x+2)^2=16 (x+2)= 4 or -4 X= 6 or -6 is my answer
Is that correct @UskiDoll
\[x^2+4x-12=0 \] \[x^2+4x=12 \] \[4 \times \frac{1}{2} = \frac{4}{2} = 2\] \[2^2 = 4\] \[x^2+4x+4=12+4 \] \[x^2+4x+4=16 \] \[(x+2)^2 = 16 \] square root on both sides \[\sqrt{(x+2)^2} = \sqrt{16}\] \[x+2=4, x+2=-4\] \[x=2, x=-6\]
x+2 = 4 on your part is wrong. subtract 2 from both sides, but -6 is correct
Okay. Thank you
plug it in -b+_ b2-4ac \[\sqrt{b2-4ac}\] / 4a good luck
The problem is already done ^_^

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