$\left(4+2\right)^2-12$
$\left(4a-2b\right)+6a^2$
$\int_{\frac{1}{2}}^{\infty}\left(\frac{ln\left(2x\right)}{x^2}\right)dx$
$-5-\left|-3+\left(-4\right)-\left(-1\right)\right|$
$2x+3y;\:x=3;\:y=2$
$9x^2-20x-21=0$
$-18\:-\:10\:\cdot\:13$
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