$3x+8=-2x-17$

$\frac{dq}{dx}\left(q^3\right)=\left(p-2\right)^{\frac{1}{2}}\left(p^2+1\right)\left(p^4+6\right)$

$\int3\cdot\left(tan\left(4x\right)\right)^3\cdot\left(4\cdot tan\left(4x\right)^2+4\right)\cdot dx$

$3.2\:-\:\left(1.98\:+\:2.5\right)$

$\frac{4}{\left(-3.9\right)^2+2\left(-3.9\right)-8}$

$2\frac{\sqrt{6}}{\sqrt{3}}$

$2y^3-5x^2=89$