## Task B-2.1.

How many integers $a$ are there for which both solutions of the equation $(x-20)(x+17)=\frac{1}{4} a$ are positive real numbers?
## First solution.

$(x-20)(x+17)=\frac{1}{4} a \Leftrightarrow x^{2}-3 x-340-\frac{1}{4} a=0$.

Both solutions of the equation will be positive real numbers if the following conditions are met:

1. $D \geqslant 0 \Leftrightarrow 9+4\left(340+\frac{1}{4} a\right) \geqslant 0 \Rightarrow a \geqslant-1369$.

2 points
2. $x_{1}+x_{2}>0$, and since $x_{1}+x_{2}=3$ this holds for any real number $a$.

1 point
3. $x_{1} x_{2}>0 \Rightarrow-340-\frac{1}{4} a>0 \Rightarrow a<-1360$.

Thus, for $a \in[-1369,-1360\rangle$ both solutions of the given equation are positive real numbers, and there are 9 integers in this interval.