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x^2-7x+12=0 solve, completing the square
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3,4
can u show how?
(x-4)(x+3)=0
\[ x^2-7x+12=0 \Rightarrow (x-4)(x+3)=0 \Rightarrow x =3,4\]
Right answer but not by completing the square: \[(x-7/2) ^{2} - 1/4 = 0\] (x-7/2) ^{2} = 1/4 (x-7/2) = sqrt (1/4) = \[\pm1/2\] x = 7/2 \[\pm\]1/2 x = 6/2 = 3 or x = 8/2 = 4
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