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What are the solutions? 1/2x^2+2x+3=0
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Well we could use quadratic formula. Using\[ax^2+bx+c\] as a format, a = .5, b=2, and c=3. \[-b \pm \sqrt{\frac{ b^2-4ac }{ 2a }}\] \[-2 \pm \sqrt{\frac{ 2^2-4(1/2)(3) }{ 2(1/2) }}\] \[-2 \pm \sqrt{\frac{ -2 }{ 1 }}\] So, it would be\[x=-2-\sqrt{-2}\] and \[-2+\sqrt{-2}\] If you feel like simplifying it a little more, we could use the imaginary number i (which equals the square root of -1) and then your solutions would be \[x=-2-2i\] or \[x=-2+2i\]
i agree with @halorazer ...
Great explanation for me too
I aim to please. ^_^
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