(B+2)(b-3)=-6 solve for b
Hi Lila;
(B+2)(b-3)=-6
I am assuming that B and b are the same and this is just a typographical error.
(b+2)(b-3)=-6
Let's FOIL...
FIRST...(b)(b)=b^{2}
OUTER...(b)(-3)=-3b
INNER...(2)(b)=2b
LAST...(2)(-3)=-6
b^{2}-3b+2b-6=-6
b^{2}-b-6=-6
Let's add 6 to both sides...
b^{2}-b-6+6=-6+6
b^{2}-b=0
b(b-1)=0
b=0, and/or (b-1)=0, b=1.
Let's check our work...
b=0
(b+2)(b-3)=-6
(0+2)(0-3)=-6
(2)(-3)=-6
-6=-6
b=1
(1+2)(1-3)=-6
(3)(-2)=-6
-6=-6