1 math Q

[n(n-1)][(n-2)(n+1)]=840
(n^2-n)(n^2-n-2)=840
let y=n^2-n
y(y-2)=840
Solving the quadratic equation , gives y= -28 or 30
therefore, -28=n^2-n or 30=n^2-n
Solving the above quadratic equation, gives
n=[1+(-111)^(1/2)]/2 or n=[1-(-111)^(1/2)]/2 or n= -5 or n=6

For solving the quadratic equation ,
ax^2+bx+c=0 => x=[-b+(b^2-4ac)^(1/2)]/(2a) or x=[-b-(b^2-4ac)^(1/2)]/(2a)

840 = 2*2*2*3*5*7

因此840 = 4*5*6*7

得n = 6

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