$x\sqrt{-252}$
$13=\frac{x}{3}-4$
$-6x^2-11x-6=0$
$2.\left(x\:+\:2\right)\left(x\:+\:4\right)$
$\int_{-2}^{\infty}\left(\frac{1}{x^2+4x+9}\right)dx$
$\left(\frac{3}{4}r^5-\frac{1}{2}p^4\right)^5$
$\frac{\left(5x+4\right)}{\left(5x-4\right)}+\frac{\left(5x-4\right)}{5x+4}=\frac{13}{6}$
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