$4x^2+z^2+4z+4y=0$
$x^2\:+\:7x\:-\:30$
$\frac{x^2+7x+10}{x+3}$
$\left(3a^x+2b^2\right)^3$
$\frac{dy}{dx}\left(8x^5+3y^5=11xy\right)$
$\frac{\sec\left(x\right)\cos\left(x\right)}{\csc^2\left(x\right)}=\sin^2\left(x\right)$
$\frac{x+3}{-4}=\frac{x-5}{4}$
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