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solve 2x^2-4x+7=0 by using the quadratic formula
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\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with \[a=2,b=-4,c=7\] gives \[\frac{4\pm\sqrt{(-4)^2-4\times 2\times 7}}{2\times 2}\] it is arithmetic from there on in
@satellite73 what would be the answer
you only need to do the arithmetic under the radical what do you get ?
so 2x2=4? @satellite73
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yes in the denominator
how about \((-4)^2-4\times 2\times 7\)?
6
@satellite73
is it a,b,c? @satellite73
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check the pic i posted and whats the answer there
I think, that is the answer
thanks
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