Simplify the equation. Then solve by the quadratic formula..

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Simplify the equation. Then solve by the quadratic formula..

Mathematics
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3x^2-4x+3=0
\[3x^2+(-4)x+3=0 \\ \text{ compare this to } \\ ax^2+bx+c=0\] what does a,b,c have to equal ?
x= -(-4)+- square root (-4)^2 -4(3)(3)/2(3)

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\[a=3,b=-4,c=3 \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-(-4) \pm \sqrt{(-4)^2-4(3)(3)}}{2(3)}\] is correct so far
-(-4)=? (-4)^2=? 4(3)(3)=? 2(3)=?
4+square root 16-36/6 4+- square root -20/6
Please dont talk down to me. Im in college. Im just trying to figure out where I went wrong.
\[x=\frac{4 \pm \sqrt{16-36}}{6} \\ x=\frac{4 \pm \sqrt{-20}}{6}\] that is also correct so far
who is talking down to you?
Any question I ask on the math group does..
my example says to do this--
I don't know where I was talking down to you at but sorry . I will stop trying to help.
6(4+-4i square root)/6 cancel out the 6
Sorry I wasnt saying you were I was say please dont do what others have
6(4+-4i square root 5)/6 cancel out the 6
My final answer is 4+-2i square root 5 but its wrong

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