(y-4)(4y^2-3y-5)=0

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Solution for (y-4)(4y^2-3y-5)=0 equation: 


Simplifying
(y + -4)(4y2 + -3y + -5) = 0

Reorder the terms:
(-4 + y)(4y2 + -3y + -5) = 0

Reorder the terms:
(-4 + y)(-5 + -3y + 4y2) = 0

Multiply (-4 + y) * (-5 + -3y + 4y2)
(-4(-5 + -3y + 4y2) + y(-5 + -3y + 4y2)) = 0
((-5 * -4 + -3y * -4 + 4y2 * -4) + y(-5 + -3y + 4y2)) = 0
((20 + 12y + -16y2) + y(-5 + -3y + 4y2)) = 0
(20 + 12y + -16y2 + (-5 * y + -3y * y + 4y2 * y)) = 0
(20 + 12y + -16y2 + (-5y + -3y2 + 4y3)) = 0

Reorder the terms:
(20 + 12y + -5y + -16y2 + -3y2 + 4y3) = 0

Combine like terms: 12y + -5y = 7y
(20 + 7y + -16y2 + -3y2 + 4y3) = 0

Combine like terms: -16y2 + -3y2 = -19y2
(20 + 7y + -19y2 + 4y3) = 0

Solving
20 + 7y + -19y2 + 4y3 = 0

Solving for variable 'y'.

The solution to this equation could not be determined.

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