$\frac{1}{3}\left(y+12\right)-12$
$12j+2j-4+3j^2$
$a+2\:-3\left\{\right\}$
$\left(y+4\right)\left(y-1\right)$
$\frac{3x^2}{2x^5}$
$7\left(-5\right)^2+21\left(-5\right)-28$
$16x^2-25=0$
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