The sum of the real roots of the equation $x^4-2 x^3+x-380=0$ is

Considering the question to be, to find the roots of the equation $x^4-2 x^3+x-380=0$ Now, by trial error method we find that $x=5$ is a solution to the equation $\Rightarrow \quad 5^4-2 \times 5^3+5-380=0$ $\begin{aligned} & =625-250+5-380=0 \\ & =380-380=0\end{aligned}$ Thus, $x=5$ is a solution. We also find that $x=-4$ is a solution $\begin{aligned} & (-4)^4-2 \times(-4)^3-4-380=0 \\ & =256+128-4-380 \\ & =380-380=0\end{aligned}$ The real roots are 5 and -4 . Sum of real roots are $5+(-4)=1$