(x^2+7x+12)-(x+3)(x+4)=1

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Solution for (x^2+7x+12)-(x+3)(x+4)=1 equation: 


Simplifying
(x2 + 7x + 12) + -1(x + 3)(x + 4) = 1

Reorder the terms:
(12 + 7x + x2) + -1(x + 3)(x + 4) = 1

Remove parenthesis around (12 + 7x + x2)
12 + 7x + x2 + -1(x + 3)(x + 4) = 1

Reorder the terms:
12 + 7x + x2 + -1(3 + x)(x + 4) = 1

Reorder the terms:
12 + 7x + x2 + -1(3 + x)(4 + x) = 1

Multiply (3 + x) * (4 + x)
12 + 7x + x2 + -1(3(4 + x) + x(4 + x)) = 1
12 + 7x + x2 + -1((4 * 3 + x * 3) + x(4 + x)) = 1
12 + 7x + x2 + -1((12 + 3x) + x(4 + x)) = 1
12 + 7x + x2 + -1(12 + 3x + (4 * x + x * x)) = 1
12 + 7x + x2 + -1(12 + 3x + (4x + x2)) = 1

Combine like terms: 3x + 4x = 7x
12 + 7x + x2 + -1(12 + 7x + x2) = 1
12 + 7x + x2 + (12 * -1 + 7x * -1 + x2 * -1) = 1
12 + 7x + x2 + (-12 + -7x + -1x2) = 1

Reorder the terms:
12 + -12 + 7x + -7x + x2 + -1x2 = 1

Combine like terms: 12 + -12 = 0
0 + 7x + -7x + x2 + -1x2 = 1
7x + -7x + x2 + -1x2 = 1

Combine like terms: 7x + -7x = 0
0 + x2 + -1x2 = 1
x2 + -1x2 = 1

Combine like terms: x2 + -1x2 = 0
0 = 1

Solving
0 = 1

Couldn't find a variable to solve for.

This equation is invalid, the left and right sides are not equal, therefore there is no solution.

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