(8x+1)(4x^2-19x-5)=0

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Solution for (8x+1)(4x^2-19x-5)=0 equation: 


Simplifying
(8x + 1)(4x2 + -19x + -5) = 0

Reorder the terms:
(1 + 8x)(4x2 + -19x + -5) = 0

Reorder the terms:
(1 + 8x)(-5 + -19x + 4x2) = 0

Multiply (1 + 8x) * (-5 + -19x + 4x2)
(1(-5 + -19x + 4x2) + 8x * (-5 + -19x + 4x2)) = 0
((-5 * 1 + -19x * 1 + 4x2 * 1) + 8x * (-5 + -19x + 4x2)) = 0
((-5 + -19x + 4x2) + 8x * (-5 + -19x + 4x2)) = 0
(-5 + -19x + 4x2 + (-5 * 8x + -19x * 8x + 4x2 * 8x)) = 0
(-5 + -19x + 4x2 + (-40x + -152x2 + 32x3)) = 0

Reorder the terms:
(-5 + -19x + -40x + 4x2 + -152x2 + 32x3) = 0

Combine like terms: -19x + -40x = -59x
(-5 + -59x + 4x2 + -152x2 + 32x3) = 0

Combine like terms: 4x2 + -152x2 = -148x2
(-5 + -59x + -148x2 + 32x3) = 0

Solving
-5 + -59x + -148x2 + 32x3 = 0

Solving for variable 'x'.

The solution to this equation could not be determined.

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