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-3x^2+(1/2)x=2(1/2)
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isn't 2(1/2) == 1??
no its 2.5 sorry
\[-3x^2+(1/2)x=1\] \[-3x^2+(1/2)x-1=0\]
ok put 2.5 in place of my 1. Use the quadratic formula to solve it
\[-3x^2+\frac12x=2\frac12\]\[\implies 3x^2-\frac 12x+\frac52=0\]\[\implies 6x^2-1x+5=0\]
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i have to have a imaginary number
\[\huge x= {-b \pm \sqrt{b^2-4ac} \over 2a}\]
\[x={1 \pm \sqrt{1-4\cdot30} \over 2\cdot6}={1 \pm \sqrt{119}i \over 12}\]
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