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bobsmithBest ResponseYou've already chosen the best response.0
quadratic formula In: \[ax^2 + bx + c = 0\] \[x _{1,2} = (b \pm \sqrt{b^24ac}) / 2a\] plug it in \[x _{1,2} = ((2) \pm \sqrt{(2)^24(1)(4)}) / 2(1)\] which equals 2 plus or minus sqrt(416) = 2+2(i)sqrt(3) or 22(i)sqrt(3) \[i = \sqrt{1}\]
 2 years ago
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