$3x^2-19x+6=0$

$\left(x+1\right)\cdot\left(x-1\right)+4$

$\frac{x^4}{4x^3}$

$\int\left(\left(2x^3+1\right)^7x^2\right)dx$

$-4\cdot\left(6-\left(+4\right)\right)\:-\frac{\left(\left(-8\right)\:-\left(+16\right)\right)}{2}$

$4x^2+3y^2-16x+18y=-31$

$3\left(2\right)^3+6\left(2\right)^2-\left(2\right)+3$