$-10\left(10+14\right)+7\left(15+6\right)-12\left(3-3\right)$
$\int\frac{\left(4x^2+8\right)}{x^4+4x^2}dx$
$\left(x+y\right)^2+3x=15$
$5\left(x-3\right)=-5$
$x^2+24+10x$
$-189-243$
$-56x-35x^2$
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