Знайдіть суму коренів рівняння x(x-3)(x-2)(x-1)=24

$x(x-3)(x-2)(x-1)=24\\\\ \ \ [x(x-3)]*[(x-2)(x-1)]=24\\\\ \ \ [x^2-3x]*[x^2-2x-x+2]=24\\\\ (x^2-3x)*(x^2-3x+2)=24\\\\ x^2-3x=t\\\\ t*(t+2)=24\\\\ t^2+2t-24=0\\\\ t^2+6t-4t-24=0\\\\ t(t+6)-4(t+6)=0\\\\ (t-4)(t+6)=0\\\\ t-4=0\ \ or\ \ t+6=0\\\\ x^2-3x-4=0\ \ or\ \ x^2-3x+6=0\\\\ x^2-4x+x-4=0\ \ or\ \ x^2-2*x*1.5+1.5^2-1.5^2+6=0\\\\ x(x-4)+(x-4)=0\ \ or\ \ (x-1.5)^2-2.25+6=0\\\\ (x+1)(x-4)=0\ \ or\ \ (x-1.5)^2+3.75=0\\\\ x+1=0\ \ or\ \ x-4=0\\\\ x=-1\ \ or\ \ x=4$

Ответ: $-1;\ 4$