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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

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

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27x-3=105 27x =105+3 27x = 108 27x/27=108/27 x= 4

27x-3=105 27x=105+3 27x=108 x=108/27 x=4ans

7x+2-2=16-27x = 147x/7 = 14/7x = 2

2275.91 = 28xx = 81.2825

Is the same as: 27x+1 = 3 Deduct 1 from both sides: 27x+1-1 = 3-1 27x = 2 Divide both sides by 27: x = 2/27

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27x-3=105 27x =105+3 27x = 108 27x/27=108/27 x= 4

27x-3=105 27x=105+3 27x=108 x=108/27 x=4ans

7x+2-2=16-27x = 147x/7 = 14/7x = 2

2275.91 = 28xx = 81.2825

20x2 - 27x -8 20 * -8 is -160, and -32 and 5 add to -27 20x2 - 32x + 5x - 8 4x(5x-8) + 1(5x-8) (4x + 1)(5x-8)

9 + 9y = 27x -9 -9 9y = 27x - 9 9y/9 = 27x/9 - 9/9 y = 3x -1

It is: 23-27x when simplified

Is the same as: 27x+1 = 3 Deduct 1 from both sides: 27x+1-1 = 3-1 27x = 2 Divide both sides by 27: x = 2/27

3x2 + 27x +60

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.