(7x+2)(4x^2-19x-5)=0

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Solution for (7x+2)(4x^2-19x-5)=0 equation: 


Simplifying
(7x + 2)(4x2 + -19x + -5) = 0

Reorder the terms:
(2 + 7x)(4x2 + -19x + -5) = 0

Reorder the terms:
(2 + 7x)(-5 + -19x + 4x2) = 0

Multiply (2 + 7x) * (-5 + -19x + 4x2)
(2(-5 + -19x + 4x2) + 7x * (-5 + -19x + 4x2)) = 0
((-5 * 2 + -19x * 2 + 4x2 * 2) + 7x * (-5 + -19x + 4x2)) = 0
((-10 + -38x + 8x2) + 7x * (-5 + -19x + 4x2)) = 0
(-10 + -38x + 8x2 + (-5 * 7x + -19x * 7x + 4x2 * 7x)) = 0
(-10 + -38x + 8x2 + (-35x + -133x2 + 28x3)) = 0

Reorder the terms:
(-10 + -38x + -35x + 8x2 + -133x2 + 28x3) = 0

Combine like terms: -38x + -35x = -73x
(-10 + -73x + 8x2 + -133x2 + 28x3) = 0

Combine like terms: 8x2 + -133x2 = -125x2
(-10 + -73x + -125x2 + 28x3) = 0

Solving
-10 + -73x + -125x2 + 28x3 = 0

Solving for variable 'x'.

The solution to this equation could not be determined.

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