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2x(x-2)+4<3(x-1)+5
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\[2x(x-2)+4<3(x-1)+5\]
2x(x-2)+4<3(x-1)+5 2x^2 -4x+4<3x-3+5 2x^2 -7x +2
<0*
right?
\[2x^2 - 4x + 4 < 3x-3+5\]
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Now take it to the completed square form
\[2x^2 - 7x +2 < 0\]
use \[D= \frac{\pm b \sqrt{b^2-4ac} }{ 2a } \]
where a = 2 b = -7 c = 2
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