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a = 1, b = 6, c = 5 Solve using the Quadratic Formula
ANS: -5 or -1 -(2/5) or -(1/2) -(2/5) or -(1/3) 5 or 1
okay and you already know the quadratic formula. \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

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\[\frac{ -6 \pm \sqrt{6^2 - 20} }{ 2}\] This would be a simpler form of the formula given the terms. I just eliminated the a = 1 and added the -4(5).
yep. now square the 6 under the radical
I'm sure you know what \(6^2\)is right?
\[\frac{ -6 \pm 3.5 }{ 2 }\]
The problem being, when I solve, I get 4.75 and -1.25, which if you look at the answers above, these are not it.
okay, first off, how did you get 3.5?
\(6^2=36\) so you would take \(\large\sqrt{36-20}\). What is 36-20?
\[\frac{ -6 \pm \sqrt{36 - 20}}{ 2} = \frac{ -6 \pm \sqrt{12} }{ 2 }\] \[\sqrt{12} = 3.5\]
36-20 does not equal 12. double check your math there
\[\frac{ -6 \pm 16 }{ 2} = 5 \space or \space -10\]
ANS: -5 or -1 -(2/5) or -(1/2) -(2/5) or -(1/3) 5 or 1
you forgot something. \(\large\sqrt{16}=?\)
because you have \(\LARGE \frac{-6\pm\sqrt{16}}{2}\)
right. Even if you take 4 \[\frac{ -6 \pm 4 }{ 2 } = -2/2 = -1 \space or -5\]
there it is. I was a mess.
I'm giving the computer back to my girlfriend. Thanks!
hahaha it's alright :) do you need any other help? I'll be happy to help. Sometimes you just have to slow down a bit and think xp
you're welcome! :D you know where to find me if needed.

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