$3x^2+12x+6$
$x^2+2x-3\ge0$
$18\left(.30+15-\left(-6\right)\right)$
$\left(\frac{x}{3x-1}\right)^2-4\left(\frac{x}{3x-1}\right)$
$12x\:-\:4\left(2+3x\right)$
$-\left|8\right|-\left|4\right|$
$15+2\cdot\left|8-3\cdot5\right|-20$
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