6.180. $\left(x^{2}+x+1\right)+\left(x^{2}+2 x+3\right)+\left(x^{2}+3 x+5\right)+\ldots$

$$
\ldots+\left(x^{2}+20 x+39\right)=4500
$$
## Solution.

The number of addends is 20. Write the equation in the form

$$
\begin{aligned}
& \left(x^{2}+x^{2}+\ldots+x^{2}\right)+(x+2 x+3 x+\ldots+20 x)+(1+3+5+\ldots+39)=4500 \Leftrightarrow \\
& 20 x^{2}+x(1+2+3+\ldots+20)+(1+3+5+\ldots+39)=4500
\end{aligned}
$$

By the formula for the sum of the terms of an arithmetic progression, we have

$20 x^{2}+x \cdot \frac{1+20}{2} \cdot 20+\frac{1+39}{2} \cdot 20=4500 \Leftrightarrow 2 x^{2}+21 x-410=0$, from which $x_{1}=-\frac{41}{2}, x_{2}=10$.

Answer: $x_{1}=-\frac{41}{2}, x_{2}=10$.