2. Let the equation be $x^{2}-402 x+k=0$ where one root plus 3 is 80 times the other root. Then $k=$ $\qquad$
2. 1985

Let the roots of the equation be $x_{1}, x_{2}$. From the problem, we have
$$
x_{1}+3=80 x_{2} \text {, i.e., } x_{1}=80 x_{2}-3 \text {, }
$$

Also, $x_{1}+x_{2}=402$,
$$
\therefore 80 x_{2}-3+x_{2}=402 \text {, }
$$

Thus, $x_{2}=5, x_{1}=80 x_{2}-3=397$
Therefore, $k=x_{1} x_{2}=397 \times 5=1985$,