Can someone help explain how to do quadratic equations with the quadratic formula? I'll give you a problem. See comments.

- poopsiedoodle

- schrodinger

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- poopsiedoodle

\[\huge 0 = -4(-4t^2 -8t -1)\]

- hba

Well the first step is to compare your equation with the standard form and get a,b and c
\[\huge\ ax^2+bx+c=0\]

- poopsiedoodle

So, how would I do that with this one? Multiply the parentheses stuff by -4?

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## More answers

- hba

Seeing your question i would suggest you to first divide your whole equation by -4 and then compare

- poopsiedoodle

\[\Huge 0=-4t^2 -8t-1\]

- hba

What is a,b and c in this ?

- poopsiedoodle

a=-4 b=-8 c=-1

- hba

Good,Now you can use the quad formula.
\[\huge\ x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

- hba

The other think which you can do with your equation is that you can take -1 common and then compare so everything becomes positive.you can also do it straight away :)

- poopsiedoodle

so, t = -8*√64-4*-4
------------------
2*-4 ?

- poopsiedoodle

which would then be t=-8*(8+4)
-----------
-8

- poopsiedoodle

am I getting it right?

- hba

It should be +- not *

- poopsiedoodle

I don't get it... wouldn't you be multiplying -8 and \[\pm √64-4*-4\]?

- hba

t=-8(+-)(8+4)
-----------
-8

- poopsiedoodle

Now I'm rather confused.

- hba

Lemme show you.

- hba

I appreciate you working with me though.

- poopsiedoodle

I guess I could do that \(\Huge\mathsf{:P}\)

- hba

a=-4 b=-8 c=-1
\[\huge\ t=\frac{ -(-8) \pm \sqrt{(-8)^2-4(-4)(-1)} }{ 2(-4) }\]

- poopsiedoodle

Ok, but what does the \[\pm\] do? what's its significance?

- hba

Well when we simplify it we would be taking the adding part on one side and subracting part on the other side.

- poopsiedoodle

so it'd be 8+√blahblahblah
---------------
2-(-4)?

- hba

Either,
8+√blahblahblah
---------------
2-(-4)?
Or,
8-√blahblahblah
---------------
2-(-4)?

- poopsiedoodle

so the \[\pm\] just breaks up between the denominator and the numerator?

- hba

Exactly.

- poopsiedoodle

Ah. I guess I'll try another problem here and ask you if it's right or something.

- hba

Sure :)

- poopsiedoodle

\[\huge 0=-2x ^{2} +14x +20\]
\[\large x=\frac{ -14\pm \sqrt{14^2 - 4(-2+20)} }{ -4 }\]
\[\large x= \frac{ -14+14-72 }{ -4 }\]

- poopsiedoodle

and the answer is
\[\huge \frac{ -7\pm \sqrt{89} }{ 2 }\]. Would I be right?

- hba

Its is -4(a)(c) not -4(a+c)

- poopsiedoodle

oh. Well that ruins that then.

- hba

Yes it does :)
Do it on a paper and tell me what you got from that equation :)

- poopsiedoodle

\[\Large x=\frac{ -14\pm \sqrt{14^2 - 4(-2*20)} }{ -4 }\]
\[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\]
\[\Large x=\frac{ -14\pm 14-160} { -4}\]
\[\Large x=\frac{ -14\pm -146} {-4}\]
set x to 0 and divide all by -2 to get \[\Large \frac{ 7\pm 73} { 2}\]

- poopsiedoodle

and now I'm confused.

- poopsiedoodle

oh wait, I forgot to keep the x's in there. Crap.

- poopsiedoodle

Think you could help me out here? My mind is about to implode.

- hba

x=7+73/2,x=7-73/2

- hba

Btw,
(-) *(-)=+

- hba

So just because of signs it goes wrong.

- hba

\[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\]
Continue from here.

- poopsiedoodle

But I left the xs out of my work. ._.

- hba

No worries about that :)

- hba

Continue working from here.

- poopsiedoodle

\[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\]
\[\Large x=\frac{ -14\pm 14+160} {-4}\]
\[\Large x=\frac{ -14\pm 174} {-4}\]
\[\Large x=\frac {\frac{ -14\pm 174} {-4}} {-2}\]
\[\Large x=\frac {7\pm -87} {2}\]

- poopsiedoodle

And now what do I do?

- hba

I don't know what you are upto ,where did that divide by -2 come in between ?

- poopsiedoodle

Well, I'm not sure. I just thought I had to do that. I'm rather confused ._.

- hba

Hmm,If OS would be working on chrome i would have worked one problem for you.
Can we do this later ?

- poopsiedoodle

I'm using chrome right now o_o But, I guess that'd be OK :P

- hba

Thanks.

- poopsiedoodle

Thanks to you too :3

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