10.330. The lengths of the legs of a certain right triangle are the roots of the equation $a x^{2}+b x+c=0$. Find the radius of the circle inscribed in this triangle.
## Solution.

Without loss of generality, we can assume that $a>0$ (the case $a<0$ is considered analogously). Then its hypotenuse is $\sqrt{m^{2}+n^{2}}$, and the radius of the inscribed circle is $r=\frac{m+n-\sqrt{m^{2}+n^{2}}}{2}$. By the condition,

$$
\begin{aligned}
& m+n=-\frac{b}{a} ; m n=\frac{c}{a} \cdot r=\frac{m+n-\sqrt{(m+n)^{2}-2 m n}}{2}=\frac{-\frac{b}{a}-\sqrt{\frac{b^{2}}{a^{2}}-\frac{2 c}{a}}}{2}= \\
& =-\frac{b+\sqrt{b^{2}-2 a c}}{2 a} . \\
& \text { Answer: }-\frac{b+\sqrt{b^{2}-2 a c}}{2 a} .
\end{aligned}
$$