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

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


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

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

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

Multiply (4 + 5x) * (5 + -7x + 2x2)
(4(5 + -7x + 2x2) + 5x * (5 + -7x + 2x2)) = 0
((5 * 4 + -7x * 4 + 2x2 * 4) + 5x * (5 + -7x + 2x2)) = 0
((20 + -28x + 8x2) + 5x * (5 + -7x + 2x2)) = 0
(20 + -28x + 8x2 + (5 * 5x + -7x * 5x + 2x2 * 5x)) = 0
(20 + -28x + 8x2 + (25x + -35x2 + 10x3)) = 0

Reorder the terms:
(20 + -28x + 25x + 8x2 + -35x2 + 10x3) = 0

Combine like terms: -28x + 25x = -3x
(20 + -3x + 8x2 + -35x2 + 10x3) = 0

Combine like terms: 8x2 + -35x2 = -27x2
(20 + -3x + -27x2 + 10x3) = 0

Solving
20 + -3x + -27x2 + 10x3 = 0

Solving for variable 'x'.

The solution to this equation could not be determined.

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