I told you that formula is wrong.Quote:

Originally Posted bydanielwu

If that is exactly what your book says then it wrong and you should burn it to death.

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- Sep 8th 2006, 09:22 AMThePerfectHackerQuote:

Originally Posted by**danielwu**

If that is exactly what your book says then it wrong and you should burn it to death. - Sep 8th 2006, 09:30 AMdanielwuQuote:

Originally Posted by**ThePerfectHacker**

$\displaystyle

4h^2-8rh+b^2=0

$

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

$\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \frac{8\pm \sqrt{64 - 16}}{8}$

Am i right? pls correct me if i am wrong. thanks guys - Sep 8th 2006, 09:30 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

while still alive, be disembowelled, have its entrails burnt before its eyes, the

be hung on London Bridge until it smells so badly that the Aldermen demand

that it be taken down and disposed of somewhere where nobody will notice

(Utah?).

RonL - Sep 8th 2006, 09:33 AMThePerfectHackerQuote:

Originally Posted by**danielwu**

For the third time the formula is,

$\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ - Sep 8th 2006, 09:33 AMCaptainBlackQuote:

Originally Posted by**danielwu**

**Im**PerfectHacker is composing a blistering reply to this

so I will just say, people have told you repeatedly that this is not the

quadratic formula.

RonL - Sep 8th 2006, 09:35 AMThePerfectHackerQuote:

Originally Posted by**CaptainBlank**

- Sep 8th 2006, 09:39 AMThePerfectHacker
Your new updated post is now correct.

- Sep 8th 2006, 10:19 AMtopsquarkQuote:

Originally Posted by**CaptainBlack**

-Dan - Sep 8th 2006, 12:05 PMdan
reading this thread gave me the best laugh i've had in a while...

:D ^_^

$\displaystyle \sqrt[iel]{~dan}$