16+(x-4)(x+4)-x^2=0

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Solution for 16+(x-4)(x+4)-x^2=0 equation: 


Simplifying
16 + (x + -4)(x + 4) + -1x2 = 0

Reorder the terms:
16 + (-4 + x)(x + 4) + -1x2 = 0

Reorder the terms:
16 + (-4 + x)(4 + x) + -1x2 = 0

Multiply (-4 + x) * (4 + x)
16 + (-4(4 + x) + x(4 + x)) + -1x2 = 0
16 + ((4 * -4 + x * -4) + x(4 + x)) + -1x2 = 0
16 + ((-16 + -4x) + x(4 + x)) + -1x2 = 0
16 + (-16 + -4x + (4 * x + x * x)) + -1x2 = 0
16 + (-16 + -4x + (4x + x2)) + -1x2 = 0

Combine like terms: -4x + 4x = 0
16 + (-16 + 0 + x2) + -1x2 = 0
16 + (-16 + x2) + -1x2 = 0

Combine like terms: 16 + -16 = 0
0 + x2 + -1x2 = 0
x2 + -1x2 = 0

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

Solving
0 = 0

Couldn't find a variable to solve for.

This equation is an identity, all real numbers are solutions.

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