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se the quadratic formula to solve...2x^2-x=-2. I get halfway through and lose the answer it sems
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\[2x^2-x+2=0\] is a start, then use \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with \[a=2,b=-1,c=2\] unless you think this one factors easily (i don't)
you get \[\frac{1\pm\sqrt{1-4\times 2\times 2}}{2\times 2}=\frac{1\pm\sqrt{-15}}{4}\]
It doesn't, I spent forever on this
As far as factoring easily
which you can then write as \[\frac{1}{4}-\frac{\sqrt{15}}{4}i\]and its conjugate \[\frac{1}{4}+\frac{\sqrt{15}}{4}i\]
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