1b^2+1b=2

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

Simplifying
1b2 + 1b = 2

Reorder the terms:
1b + 1b2 = 2

Solving
1b + 1b2 = 2

Solving for variable 'b'.

Reorder the terms:
-2 + 1b + 1b2 = 2 + -2

Combine like terms: 2 + -2 = 0
-2 + 1b + 1b2 = 0

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

Subproblem 1
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
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 = {-2, 1}