$\left(4x+13y\right)^2$
$x^4+3x^3-11x^2-12x+28$
$\tan^2\theta\:+\tan\theta\:-30=0$
$3x+5=20$
$\frac{\left(-16\right)\left(-2\right)\left(-4\right)}{\left(+32\right)}$
$\frac{27}{27g+18}$
$-3x\:+5\:>\:5x\:-3$
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