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20x^2 + 27x + 9 = 0

20x^2 + 15x +12x + 9 = 0

5x(4x + 3) + 3(4x + 3) = 0

(5x + 3)(4x + 3) = 0

So either (5x + 3) = 0 which yields x = -3/5 or (4x + 3) = 0 which yields x = -3/4

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Q: How do you solve 20x2 plus 27x plus 9 equals 0?

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3x2 + 27x = 30 ∴ x2 + 9x - 10 = 0 ∴ (x + 10)(x - 1) = 0 ∴ x ∈ {-10, 1}

To solve this we simply plug in the value of x where any x's are found in the equation. 1) 27x - 27; x = -9 2) 27(-9) - 27 3) -243 - 27 = -270 So, 27x - 27 = -270 if x = -9.

It depends on the value of 'x'. 'y' can have as many different values as 'x' can.

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