how do i solve this problem b^2+6-7b?

Answer 1

First you need to rewrite the trinomial to read b^2-7b+6.

Then, factor the trinomial. Find two numbers with a product of 6, and a sum of 7.

These two numbers are 6 and 1, but because the 7 in the original problem is negative, you'd need to use -6 and -1.

(b-1)(b-6)=0
b-1=0 or b-6=0

b=1 or b=6

plug the answers into the original problem, and they both work :)

sorry this wasn't explained too well :)

Answer 2

Is it equal to something or do you want to simplify it?

b^2+6-7b = b^2 -7b -6

if equal to zero, you can factor it and solve
b^2 -7b +6 = 0
(b - 6)(b - 1) = 0

solve
divide both sides by (b - 1) to get
(b-6) = 0
b -6 + 6 = 0 +6
b=6

divide both sides by (b - 6) to get
(b-1) = 0
b -1 + 1 = 0 +1
b=1

answers b=1 and b=6

checking
substitute b=1
b^2 -7b +6 = 0
1^2 -7(1) +6 = 0
1 -7 +6 = 0
0 = 0 so b=1 is a valid solution

substitute b=6
b^2 -7b +6 = 0
6^2 -7(6) +6 =0
36 - 42 +6 =0
0=0 so b=6 is also a valid solution

final answer b=1, b=6

Answer 3

rearrange

b^2 -7b + 6

factor out

(b - 6)(b - 1)

b=6, b= 1

Answer 4

b^2 + 6 - 7b = b^2 - 7b + 6
=(b -6)(b -1)

To check: expand (b-6)(b-1) using FOIL.

Answer 5

b^2- 7b + 6 = 0
1. (b-6)(b-1) = 0
2. b = 6,1