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2x^2-7x+3=0 completeing the square im using the quadratic formula but i keep getting - sqare root so do i use imaginary numbers?
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the discriminant is\[b^2-4ac=(-7)^2-4(2)(3)=25>0 \]so you have made a mistake somewhere.
There for making it -7i+_\[-7i \pm \sqrt{73}\]/4
All over 4?
show your work, you have made a mistake somewhere...
\[2x^2-7x+3=0\]\[a=2\]\[b=-7\]\[c=3\]\[x={-b \pm \sqrt{b^2-4ac} \over 2a}\]plug in the numbers and tell me what you get.
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Same answer.
-73
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