^{2}

^{2}-3b+2b-6=-6

^{2}-b-6=-6

^{2}-b-6+6=-6+6

^{2}-b=0

**b=0, and/or**(b-1)=0

**, b=1.**

(B+2)(b-3)=-6 solve for b

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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

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

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