# How do you solve (5-x)^2(x-13/2)<0?

$x \in \left(- \infty , \frac{13}{2}\right) - \left\{5\right\}$
Observe that $\forall x \in \mathbb{R} , {\left(5 - x\right)}^{2} \ge 0 ,$ equality holds $\iff x = 5.$
Hence, (5-x)^2(x-13/2)<0 iff (x-13/2)<0 & x!=5 iff x<13/2, x!=5.
This can be put as $x \in \left(- \infty , \frac{13}{2}\right) - \left\{5\right\}$