b^2+4b=5

Solution for b^2+4b=5 equation:

Simplifying
b2 + 4b = 5

Reorder the terms:
4b + b2 = 5

Solving
4b + b2 = 5

Solving for variable 'b'.

Reorder the terms:
-5 + 4b + b2 = 5 + -5

Combine like terms: 5 + -5 = 0
-5 + 4b + b2 = 0

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

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

Simplifying
1 + -1b = 0

Solving
1 + -1b = 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 + -1b = 0 + -1

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

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

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

Simplifying
b = 1
Solution
b = {-5, 1}