b(b+2)=35

Solution for b(b+2)=35 equation:

Simplifying
b(b + 2) = 35

Reorder the terms:
b(2 + b) = 35
(2 * b + b * b) = 35
(2b + b2) = 35

Solving
2b + b2 = 35

Solving for variable 'b'.

Reorder the terms:
-35 + 2b + b2 = 35 + -35

Combine like terms: 35 + -35 = 0
-35 + 2b + b2 = 0

Factor a trinomial.
(-7 + -1b)(5 + -1b) = 0

Subproblem 1
Set the factor '(-7 + -1b)' equal to zero and attempt to solve:

Simplifying
-7 + -1b = 0

Solving
-7 + -1b = 0

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

Add '7' to each side of the equation.
-7 + 7 + -1b = 0 + 7

Combine like terms: -7 + 7 = 0
0 + -1b = 0 + 7
-1b = 0 + 7

Combine like terms: 0 + 7 = 7
-1b = 7

Divide each side by '-1'.
b = -7

Simplifying
b = -7
Subproblem 2
Set the factor '(5 + -1b)' equal to zero and attempt to solve:

Simplifying
5 + -1b = 0

Solving
5 + -1b = 0

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

Add '-5' to each side of the equation.
5 + -5 + -1b = 0 + -5

Combine like terms: 5 + -5 = 0
0 + -1b = 0 + -5
-1b = 0 + -5

Combine like terms: 0 + -5 = -5
-1b = -5

Divide each side by '-1'.
b = 5

Simplifying
b = 5
Solution
b = {-7, 5}