$-36\left(-25\right)\left(-4\right)\left(3\right)\:$
$3\sin^2x+\sin x=0$
$2x^2\:+13x=7$
$2x^2+5x+3\cdot3x^2-2x+4$
$\frac{dy}{dx}+y^{2\:}-y=0$
$4\left(x+2\right)>3\left(x-5\right)$
$\frac{1}{4}\int\left(\frac{x^3}{\sqrt{16-x^2}}\right)dx$
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