Not So Hard!!!

according to the viete's formula's if $\alpha , \beta , \gamma , \epsilon$ are rots of quartic equation $ax^4+bx^3+cx^2+dx+e$ then $\alpha+\beta+\gamma+\epsilon = \frac{-b}{a}$ hence $6(\alpha+\beta+\gamma+\epsilon) = 25$

$Sum\quad of\quad roots\quad :\quad a+b+c+d\quad =\quad -\quad \frac { -25 }{ 6 } \quad =\quad \frac { 25 }{ 6 } \\ 6\quad (a+b+c+d)\quad =\quad 25$