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Can you help me see where I am stuck finding the zeros of this quadratic equation?

Mathematics
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\[3x^2 - 12x + 11 = 0\]
quadratic formula for this one use \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with \[a=3,b=-12,c=11\]
My student solutions manual says I should obtain \[(6 \pm \sqrt{3}) \div 3\]

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Other answers:

indeed you should
Let's walk through this, so I can see where I am tripping over my shoelaces.
first you get \[\frac{-12\pm\sqrt{(-12)^2-4\times 3\times 11}}{6}\]
Wait a sec. I see that I made an error in the first term under the radical. I will go back to the drawing board and re-attempt this one on my own.
By the way, how are you producing the nice looking divisor line?
\frac{a}{b} looks like \[\frac{a}{b}\]
Got it. I made simple mistakes in my substitutions, then got hung up on trying to do algebra with unduly complicated figures.
THANKS @satellite73, you got me unstuck!

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