$2x^2-3x+12=0$
$\left(3y+3y+3z\right)^3$
$\cos^2x-\sec^2x$
$8x-6=6x+4$
$y=\left(3x^4-8x^3+6x^2\right)^{-3}$
$-25\:\left[\:6\:+\:\left(-4\right)\right]\:$
$\left(\frac{2w+5}{7w-9}\right)^{\frac{1}{2}}$
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