b^2+3b+1=11

Solution for b^2+3b+1=11 equation:

Simplifying
b2 + 3b + 1 = 11

Reorder the terms:
1 + 3b + b2 = 11

Solving
1 + 3b + b2 = 11

Solving for variable 'b'.

Reorder the terms:
1 + -11 + 3b + b2 = 11 + -11

Combine like terms: 1 + -11 = -10
-10 + 3b + b2 = 11 + -11

Combine like terms: 11 + -11 = 0
-10 + 3b + b2 = 0

Factor a trinomial.
(-5 + -1b)(2 + -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 '(2 + -1b)' equal to zero and attempt to solve:

Simplifying
2 + -1b = 0

Solving
2 + -1b = 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 + -1b = 0 + -2

Combine like terms: 2 + -2 = 0
0 + -1b = 0 + -2
-1b = 0 + -2

Combine like terms: 0 + -2 = -2
-1b = -2

Divide each side by '-1'.
b = 2

Simplifying
b = 2
Solution
b = {-5, 2}