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this is different 3x2=10x+2
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subtract 10x and 2 and factor out. Then use the quadratic formula to solve for "x" when you have it in the form \(ax^2 \pm bx \pm c\) \[\frac{-b \pm \sqrt{b^2-4ac} }{ 2a }\]
@abb0t thank you!!
is this correct 5+/-2sq19 over 3
I don't know. Lol. I didn't actually do the problem.
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