(2x+1)*(2x+3)+4(2x+1)=135

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Solution for (2x+1)*(2x+3)+4(2x+1)=135 equation:


Simplifying
(2x + 1)(2x + 3) + 4(2x + 1) = 135

Reorder the terms:
(1 + 2x)(2x + 3) + 4(2x + 1) = 135

Reorder the terms:
(1 + 2x)(3 + 2x) + 4(2x + 1) = 135

Multiply (1 + 2x) * (3 + 2x)
(1(3 + 2x) + 2x * (3 + 2x)) + 4(2x + 1) = 135
((3 * 1 + 2x * 1) + 2x * (3 + 2x)) + 4(2x + 1) = 135
((3 + 2x) + 2x * (3 + 2x)) + 4(2x + 1) = 135
(3 + 2x + (3 * 2x + 2x * 2x)) + 4(2x + 1) = 135
(3 + 2x + (6x + 4x2)) + 4(2x + 1) = 135

Combine like terms: 2x + 6x = 8x
(3 + 8x + 4x2) + 4(2x + 1) = 135

Reorder the terms:
3 + 8x + 4x2 + 4(1 + 2x) = 135
3 + 8x + 4x2 + (1 * 4 + 2x * 4) = 135
3 + 8x + 4x2 + (4 + 8x) = 135

Reorder the terms:
3 + 4 + 8x + 8x + 4x2 = 135

Combine like terms: 3 + 4 = 7
7 + 8x + 8x + 4x2 = 135

Combine like terms: 8x + 8x = 16x
7 + 16x + 4x2 = 135

Solving
7 + 16x + 4x2 = 135

Solving for variable 'x'.

Reorder the terms:
7 + -135 + 16x + 4x2 = 135 + -135

Combine like terms: 7 + -135 = -128
-128 + 16x + 4x2 = 135 + -135

Combine like terms: 135 + -135 = 0
-128 + 16x + 4x2 = 0

Factor out the Greatest Common Factor (GCF), '4'.
4(-32 + 4x + x2) = 0

Factor a trinomial.
4((-8 + -1x)(4 + -1x)) = 0

Ignore the factor 4.

Subproblem 1

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

Subproblem 2

Set the factor '(4 + -1x)' equal to zero and attempt to solve: Simplifying 4 + -1x = 0 Solving 4 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-4' to each side of the equation. 4 + -4 + -1x = 0 + -4 Combine like terms: 4 + -4 = 0 0 + -1x = 0 + -4 -1x = 0 + -4 Combine like terms: 0 + -4 = -4 -1x = -4 Divide each side by '-1'. x = 4 Simplifying x = 4

Solution

x = {-8, 4}

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