$x^2+2x+2y^2-4y=0$

$y^2-10y-11$

$\left(\frac{1}{2}a^2-\frac{3}{4}\right)^2$

$\frac{dy}{dx}=\frac{x+3}{3+y}$

$\int\frac{\left(6x-7\right)}{\left(x+1\right)^3}dx$

$\sec^2x-\tan^4x$

$\left(-3\right)\left(-1\right)+\left(-1\right)\left(\left(-\sqrt{81}\right)+3\left(7-4-3\left(5-\sqrt{25}+1\right)\right)\right)-\left(-16:8:\left(-2\right)\right)$