(2x+3)(4x^2-19x-5)=0

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Solution for (2x+3)(4x^2-19x-5)=0 equation: 


Simplifying
(2x + 3)(4x2 + -19x + -5) = 0

Reorder the terms:
(3 + 2x)(4x2 + -19x + -5) = 0

Reorder the terms:
(3 + 2x)(-5 + -19x + 4x2) = 0

Multiply (3 + 2x) * (-5 + -19x + 4x2)
(3(-5 + -19x + 4x2) + 2x * (-5 + -19x + 4x2)) = 0
((-5 * 3 + -19x * 3 + 4x2 * 3) + 2x * (-5 + -19x + 4x2)) = 0
((-15 + -57x + 12x2) + 2x * (-5 + -19x + 4x2)) = 0
(-15 + -57x + 12x2 + (-5 * 2x + -19x * 2x + 4x2 * 2x)) = 0
(-15 + -57x + 12x2 + (-10x + -38x2 + 8x3)) = 0

Reorder the terms:
(-15 + -57x + -10x + 12x2 + -38x2 + 8x3) = 0

Combine like terms: -57x + -10x = -67x
(-15 + -67x + 12x2 + -38x2 + 8x3) = 0

Combine like terms: 12x2 + -38x2 = -26x2
(-15 + -67x + -26x2 + 8x3) = 0

Solving
-15 + -67x + -26x2 + 8x3 = 0

Solving for variable 'x'.

The solution to this equation could not be determined.

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