$2\tan^2a+\tan^2a+1=4$

$\int_4^{\infty}\left(\frac{6cos\left(\frac{\pi}{x}\right)}{x^2}\right)dx$

$\int\left(\tan^2\left(3x\right)\sec^2\left(3x\right)\right)dx$

$\sqrt[3]{2x}-\sqrt[3]{-2}$

$\left(-3\right)^4-\left(-2\right)^3$

$\left(6x+12\right)\left(6x-18\right)$

$8x^2+6x=0$