In a ${\Delta }^{le}PQR,\angle R=\frac{\pi }{2}$  if $\tan \left(\frac{P}{2}\right)$ and $\tan \left(\frac{Q}{2}\right)$  are the roots of the equation $a{x}^2+bx+c=0\left(a\ne 0\right)$  then

$\tan \frac{P}{2}\tan \frac{Q}{2}=\frac{c}{a}$

$P+Q=90°$

$\frac{P}{2}+\frac{Q}{2}=45°$

$\tan \left(\frac{P}{2}+\frac{Q}{2}\right)=1$

$\frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{\alpha }\tan \frac{Q}{\alpha }}=1$

$=\frac{-\frac{b}{a}}{1-\frac{c}{a}}=1$

$\frac{-b}{a-c}=1$

$-b=a-c$

$c=a+b$