$16b^2\:-\:9$
$42+11\:x\:2$
$\int\:\frac{7x^2+29x+12}{x\left(x^2+4x+3\right)}\:dx$
$12x^2-4x-1=0$
$-1-\left(3x^2-x+1\right)+\left(2x^3-1\right)$
$16b^2-1\:=\:0$
$64\left(5\right)^2-9$
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