(x-1)(x-3)(x-4)(x-12)+5x^2=0解方程

(x-1)(x-3)(x-4)(x-12)+5x^2=0
(x-1)(x-12)(x-3)(x-4)+5x^2=0
(x^2-13x+12)(x^2-7x+12)+5x^2=0
[(x^2-10x+12)-3x][(x^2-10x+12)+3x]+5x^2=0
(x^2-10x+12)^2-9x^2+5x^2=0
(x^2-10x+12)^2-4x^2=0
[(x^2-10x+12)-2x][(x^2-10x+12)+2x]=0
(x^2-12x+12)(x^2-8x+12)=0
(x^2-12x+12)(x-2)(x-6)=0
x1=2
x2=6
x^2-12x+12=0
x^2-12x=-12
x^2-12x+36=36-12
(x-6)^2=24
x-6=±2√6
x3=6+2√6
x4=6-2√6

2，6

x=6

这个很好做嘛！
(x-1)(x-3)(x-4)(x-12)+5x^2=0
要使一元五次方程每项都为0，即
(x-1)(x-3)(x-4)(x-12=0与5x^2=0
因为只有0+0=0
所以得
(x-1)=0
(x-3)=0
(x-4)=0
(x-12)=0
5x^2=0-这个有个符号 ^ 不懂
求出x=1.x=3.x=4，x=12，x=？

x1=6，x2=2