Find the changes in the sign of the expression $15+4x-3x^2$ and find their extreme values 

Given $15+4x-3x^2=15+9x-5x-3x^2$

$=3\left(5+3x\right)-x\left(5+3x\right)$

 $=\left(3-x\right)\left(5+3x\right)$

If $\frac{-5}{3}<x<3$ the sign $15+4x-3x^2$ is positive $\left(\because a=-3<0\right)$

If $x<\frac{-5}{3}or x>3$ the sign of $15+4x-3x^2$ is $\frac{4ac-b^2}{4a}=\frac{4\left(-3\right)\left(15\right)-16}{4\left(-3\right)}=\frac{49}{3}$

Extreme value of the expression is  $\frac{49}{3}$