$x^2+14x+5=0$
$2\left(3f+7\right)-\left(2f\right)$
$-11b^2+2b^2$
$\frac{5}{c^{-7}d^4}$
$\lim_{x\to0}\left(\frac{2\left(e^x-1\right)\left(e^x\left(x-1\right)+1\right)}{x^3}\right)$
$\lim_{x\to\infty}\:\frac{2x-6}{x^2-9}$
$x^2-8x-1=8$
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