For all numbers B and D,the operation ＃is defined by b#d=(b-2)(d+1).If (-2)#x=12,then what's the value of x and why please give your formula and explain.

For all numbers B and D,the operation ＃is defined by b#d=(b-2)(d+1).If (-2)#x=12,then what's the value of x and why please give your formula and explain.
For all numbers B and D,the operation ＃is defined by b#d=(b-2)(d+1).If (-2)#x=12,then what's the value of x
and why please give your formula and explain.

For all numbers B and D,the operation ＃is defined by b#d=(b-2)(d+1).If (-2)#x=12,then what's the value of x and why please give your formula and explain.
就是除了加减乘除,假设现在有一个新的符号#.
#的算法就是b#d=(b-2)(d+1).
所以(-2)#x=（-2-2）（x+1）=12 所以x=-4

the answer is X=-4
and what grade's question is this for?
It is really not difficult

因为（-2）#x=12
所以（-2-2）（x+1)=12
x=-4

按照给的规律
-2#x=（-2-2）(x+1)=12
所以x=-4

任意数B和D,规定#定义为B#D=(B-2)(D+1).如果(-2)#X=12,求X的值,为什么,写出计算过程并解释.
因为B#D=(B-2)(D+1)
所以(-2)#X
=(-2-2)(X+1)
=-4X-4=12
所以X=-4
Hope I can help you

solusion:
As For all numbers B and D,the operation ＃is defined by b#d=(b-2)(d+1),
so (-2)#x=(-2-2)(x+1)=12,
then -4x-4=12,
x=-4.