(b+1)(3b+2)=

Solution for (b+1)(3b+2)= equation:

Simplifying
(b + 1)(3b + 2) = 0

Reorder the terms:
(1 + b)(3b + 2) = 0

Reorder the terms:
(1 + b)(2 + 3b) = 0

Multiply (1 + b) * (2 + 3b)
(1(2 + 3b) + b(2 + 3b)) = 0
((2 * 1 + 3b * 1) + b(2 + 3b)) = 0
((2 + 3b) + b(2 + 3b)) = 0
(2 + 3b + (2 * b + 3b * b)) = 0
(2 + 3b + (2b + 3b2)) = 0

Combine like terms: 3b + 2b = 5b
(2 + 5b + 3b2) = 0

Solving
2 + 5b + 3b2 = 0

Solving for variable 'b'.

Factor a trinomial.
(1 + b)(2 + 3b) = 0

Subproblem 1
Set the factor '(1 + b)' equal to zero and attempt to solve:

Simplifying
1 + b = 0

Solving
1 + b = 0

Move all terms containing b to the left, all other terms to the right.

Add '-1' to each side of the equation.
1 + -1 + b = 0 + -1

Combine like terms: 1 + -1 = 0
0 + b = 0 + -1
b = 0 + -1

Combine like terms: 0 + -1 = -1
b = -1

Simplifying
b = -1
Subproblem 2
Set the factor '(2 + 3b)' equal to zero and attempt to solve:

Simplifying
2 + 3b = 0

Solving
2 + 3b = 0

Move all terms containing b to the left, all other terms to the right.

Add '-2' to each side of the equation.
2 + -2 + 3b = 0 + -2

Combine like terms: 2 + -2 = 0
0 + 3b = 0 + -2
3b = 0 + -2

Combine like terms: 0 + -2 = -2
3b = -2

Divide each side by '3'.
b = -0.6666666667

Simplifying
b = -0.6666666667
Solution
b = {-1, -0.6666666667}