3x^2(x-4)-4x(x-4)+(1x-4)=0

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Solution for 3x^2(x-4)-4x(x-4)+(1x-4)=0 equation: 


Simplifying
3x2(x + -4) + -4x(x + -4) + (1x + -4) = 0

Reorder the terms:
3x2(-4 + x) + -4x(x + -4) + (1x + -4) = 0
(-4 * 3x2 + x * 3x2) + -4x(x + -4) + (1x + -4) = 0
(-12x2 + 3x3) + -4x(x + -4) + (1x + -4) = 0

Reorder the terms:
-12x2 + 3x3 + -4x(-4 + x) + (1x + -4) = 0
-12x2 + 3x3 + (-4 * -4x + x * -4x) + (1x + -4) = 0
-12x2 + 3x3 + (16x + -4x2) + (1x + -4) = 0

Reorder the terms:
-12x2 + 3x3 + 16x + -4x2 + (-4 + 1x) = 0

Remove parenthesis around (-4 + 1x)
-12x2 + 3x3 + 16x + -4x2 + -4 + 1x = 0

Reorder the terms:
-4 + 16x + 1x + -12x2 + -4x2 + 3x3 = 0

Combine like terms: 16x + 1x = 17x
-4 + 17x + -12x2 + -4x2 + 3x3 = 0

Combine like terms: -12x2 + -4x2 = -16x2
-4 + 17x + -16x2 + 3x3 = 0

Solving
-4 + 17x + -16x2 + 3x3 = 0

Solving for variable 'x'.

The solution to this equation could not be determined.

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