已知|a-2|+(b+1)+c²=0,求a²-2ab-5a²+12ac+3ab-c²-8ac-2a²的值

|a-2|+(b+1)+c²=0
可知
a-2=0
b+1=0
c=0
可得a=2
b=-1
a²-2ab-5a²+12ac+3ab-c²-8ac-2a²
=a²-5a²-2a²-2ab+3ab+12ac-8ac-c²
=-6a²+ab+4ac-c²
=-6×2²+2×（-1）+4×2×0-0²
=-24-2+0-0
=-26