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Simplifying x + (x + 1) + 3 + x(x + 1) + -3[x + (x + 1)] + 12 = 0 Reorder the terms: x + (1 + x) + 3 + x(x + 1) + -3[x + (x + 1)] + 12 = 0 Remove parenthesis around (1 + x) x + 1 + x + 3 + x(x + 1) + -3[x + (x + 1)] + 12 = 0 Reorder the terms: x + 1 + x + 3 + x(1 + x) + -3[x + (x + 1)] + 12 = 0 x + 1 + x + 3 + (1 * x + x * x) + -3[x + (x + 1)] + 12 = 0 x + 1 + x + 3 + (1x + x2) + -3[x + (x + 1)] + 12 = 0 Reorder the terms: x + 1 + x + 3 + 1x + x2 + -3[x + (1 + x)] + 12 = 0 Remove parenthesis around (1 + x) x + 1 + x + 3 + 1x + x2 + -3[x + 1 + x] + 12 = 0 Reorder the terms: x + 1 + x + 3 + 1x + x2 + -3[1 + x + x] + 12 = 0 Combine like terms: x + x = 2x x + 1 + x + 3 + 1x + x2 + -3[1 + 2x] + 12 = 0 x + 1 + x + 3 + 1x + x2 + [1 * -3 + 2x * -3] + 12 = 0 x + 1 + x + 3 + 1x + x2 + [-3 + -6x] + 12 = 0 Reorder the terms: 1 + 3 + -3 + 12 + x + x + 1x + -6x + x2 = 0 Combine like terms: 1 + 3 = 4 4 + -3 + 12 + x + x + 1x + -6x + x2 = 0 Combine like terms: 4 + -3 = 1 1 + 12 + x + x + 1x + -6x + x2 = 0 Combine like terms: 1 + 12 = 13 13 + x + x + 1x + -6x + x2 = 0 Combine like terms: x + x = 2x 13 + 2x + 1x + -6x + x2 = 0 Combine like terms: 2x + 1x = 3x 13 + 3x + -6x + x2 = 0 Combine like terms: 3x + -6x = -3x 13 + -3x + x2 = 0 Solving 13 + -3x + x2 = 0 Solving for variable 'x'. Begin completing the square. Move the constant term to the right: Add '-13' to each side of the equation. 13 + -3x + -13 + x2 = 0 + -13 Reorder the terms: 13 + -13 + -3x + x2 = 0 + -13 Combine like terms: 13 + -13 = 0 0 + -3x + x2 = 0 + -13 -3x + x2 = 0 + -13 Combine like terms: 0 + -13 = -13 -3x + x2 = -13 The x term is -3x. Take half its coefficient (-1.5). Square it (2.25) and add it to both sides. Add '2.25' to each side of the equation. -3x + 2.25 + x2 = -13 + 2.25 Reorder the terms: 2.25 + -3x + x2 = -13 + 2.25 Combine like terms: -13 + 2.25 = -10.75 2.25 + -3x + x2 = -10.75 Factor a perfect square on the left side: (x + -1.5)(x + -1.5) = -10.75 Can't calculate square root of the right side. The solution to this equation could not be determined.
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