$3x^2+1=0$

$-\frac{5}{6}ab^2-\frac{1}{6}ab^2-\frac{8}{8}ab^2$

$5x-3\left(x-4\right)$

$2\left(s-4\right)=12$

$4\left(x\:+\:7\right)\:=\:2x\:+\:28\:+\:2x$

$\frac{d}{dx}\left(\left(x^2+4\right)\left(x^3+2\right)\right)$

$\left(-.3\right)\left(-1.16\right)\left(-3\right)$