(4y+1)(3y^2+2y-4)=

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


Simplifying
(4y + 1)(3y2 + 2y + -4) = 0

Reorder the terms:
(1 + 4y)(3y2 + 2y + -4) = 0

Reorder the terms:
(1 + 4y)(-4 + 2y + 3y2) = 0

Multiply (1 + 4y) * (-4 + 2y + 3y2)
(1(-4 + 2y + 3y2) + 4y * (-4 + 2y + 3y2)) = 0
((-4 * 1 + 2y * 1 + 3y2 * 1) + 4y * (-4 + 2y + 3y2)) = 0
((-4 + 2y + 3y2) + 4y * (-4 + 2y + 3y2)) = 0
(-4 + 2y + 3y2 + (-4 * 4y + 2y * 4y + 3y2 * 4y)) = 0
(-4 + 2y + 3y2 + (-16y + 8y2 + 12y3)) = 0

Reorder the terms:
(-4 + 2y + -16y + 3y2 + 8y2 + 12y3) = 0

Combine like terms: 2y + -16y = -14y
(-4 + -14y + 3y2 + 8y2 + 12y3) = 0

Combine like terms: 3y2 + 8y2 = 11y2
(-4 + -14y + 11y2 + 12y3) = 0

Solving
-4 + -14y + 11y2 + 12y3 = 0

Solving for variable 'y'.

The solution to this equation could not be determined.

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