$\frac{x}{5}-4=0$
$-2x^2-3x$
$x^2+14x-120$
$-4c\:-\:2c$
$\left(-32x^{2}+9x^{4}-8-4x^{3}+\frac{1}{3}x\right)-\left(5x^{4}-\frac{2}{3}+3x+7x^{3}\right)$
$\left(12x+4\right)\left(12x-4\right)$
$3x\:+\:11\:+\:2x\:+\:5$
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