$x^2-14x-95$
$3v^2+6z+4-vz+vz$
$\lim_{x\to\infty}\left(\frac{3}{2^x}\right)$
$40\left(-6\right).\:\left(6\right)$
$-x^2+4x-6=0$
$y=\frac{\left(3x^2+2\right)^2}{\left(x+1\right)^2\left(5x+6\right)}$
$\left(x-2\right)\left(x+2\right)\left(x^2-4\right)$
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