9(x/2 + 5/6)^4 + 14(x/2 + 5/6)^2 - 8 = 0

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To solve the equation 9(x/2 + 5/6)^4 + 14(x/2 + 5/6)^2 - 8 = 0, let's introduce a substitution to make it easier to solve.

Let y = (x/2 + 5/6)^2

Now, the equation becomes 9y^2 + 14y - 8 = 0

To solve this quadratic equation, we can use the quadratic formula:

y = [-b ± sqrt(b^2 - 4ac)] / 2a

In this case, a = 9, b = 14, and c = -8. Plugging in these values:

y = [-14 ± sqrt(14^2 - 49(-8))] / 2*9
y = [-14 ± sqrt(196 + 288)] / 18
y = [-14 ± sqrt(484)] / 18
y = [-14 ± 22] / 18

This gives two possible values for y:

y = (8/9) or y = (-4/3)

Now, remember that y = (x/2 + 5/6)^2

Substitute the values of y back into the equation gives:

1) (x/2 + 5/6)^2 = 8/9
Taking the square root of both sides gives:
x/2 + 5/6 = ±√(8/9)
x/2 + 5/6 = ±(2/3)
x/2 = -5/6 ± 2/3
x/2 = -3/2 or x/2 = 1/6
x = -3 or x = 1/3

2) (x/2 + 5/6)^2 = -4/3
This equation does not have real solutions since the square of a real number cannot be negative.

Therefore, the solutions to the given equation are x = -3 and x = 1/3.