$\left(x-4\right)\left(x+6\right)=0$
$\frac{d}{dx}\frac{1}{5}\:at\:\frac{3}{4}$
$62\:.\:-2$
$x^2y^'^'-6y=0$
$-3-2-5-3-10$
$\left(16x^2+8y\right)\left(16x^2-8y\right)$
$-5a^2\left(-a\right)$
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