$3c^2-4c-4=0$
$\frac{3}{4}\left(-12x+4\right)+\frac{1}{2}\:\left(6x\:+\:8\right)$
$x+3\le41$
$\left(m^2+n^2-5\right)\left(m^2n^2+9\right)$
$-2\cdot2+2+-2\cdot1\cdot3$
$\left(3x\right)^2.x^{-1}$
$x^2-2x=8$
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