$j^2+28j=-43$

$\lim_{x\to0}\frac{lntan2x}{\frac{1}{x}}$

$\left(-4321\right)-\left(3453\right)$

$f\left(x\right)=3x^2+3x-3$

$\frac{2}{3}ab^2+\frac{1}{5}x^2y-\frac{4}{5}ab^2-\frac{2}{15}x^2y+\frac{2}{3}x^2y-\frac{1}{3}ab^2$

$\left(\frac{2b}{3}+c\right)\left(\frac{2b}{3}-c\right)$

$\left(6b^2-8b\right)^2$