2b^2+128=32b

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Solution for 2b^2+128=32b equation:


Simplifying
2b2 + 128 = 32b

Reorder the terms:
128 + 2b2 = 32b

Solving
128 + 2b2 = 32b

Solving for variable 'b'.

Reorder the terms:
128 + -32b + 2b2 = 32b + -32b

Combine like terms: 32b + -32b = 0
128 + -32b + 2b2 = 0

Factor out the Greatest Common Factor (GCF), '2'.
2(64 + -16b + b2) = 0

Factor a trinomial.
2((8 + -1b)(8 + -1b)) = 0

Ignore the factor 2.

Subproblem 1

Set the factor '(8 + -1b)' equal to zero and attempt to solve: Simplifying 8 + -1b = 0 Solving 8 + -1b = 0 Move all terms containing b to the left, all other terms to the right. Add '-8' to each side of the equation. 8 + -8 + -1b = 0 + -8 Combine like terms: 8 + -8 = 0 0 + -1b = 0 + -8 -1b = 0 + -8 Combine like terms: 0 + -8 = -8 -1b = -8 Divide each side by '-1'. b = 8 Simplifying b = 8

Subproblem 2

Set the factor '(8 + -1b)' equal to zero and attempt to solve: Simplifying 8 + -1b = 0 Solving 8 + -1b = 0 Move all terms containing b to the left, all other terms to the right. Add '-8' to each side of the equation. 8 + -8 + -1b = 0 + -8 Combine like terms: 8 + -8 = 0 0 + -1b = 0 + -8 -1b = 0 + -8 Combine like terms: 0 + -8 = -8 -1b = -8 Divide each side by '-1'. b = 8 Simplifying b = 8

Solution

b = {8, 8}

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