You need to solve for x the inequality such that:

`(3x-5)^2 - 9/4 >= 0`

You need to expand the binomial using the following formula such that:

`(a-b)^2 = a^2-2ab+b^2`

Reasoning by analogy yields:

`(3x-5)^2 = (3x)^2 - 2*(3x)*5 + 5^2`

`(3x-5)^2 = 9x^2 - 30x + 25`

`9x^2 - 30x + 25 - 9/4 >= 0`

Bringing the terms to a common denominator yields:

`36x^2 - 120x + 100 - 9 >= 0`

`36x^2 - 120x + 91 >= 0 `

You need to use quadratic formula to find the zeroes of quadratic equation `36x^2 - 120x + 91 = 0` such that:

`x_(1,2) = (120+-sqrt(14400 - 13104))/(2*36) ` `x_(1,2) = (120+-sqrt1296)/(2*36) => x_(1,2) = (120+-36)/(2*36)`

`x_(1,2) = (4(30+-9))/(2*36) => x_(1,2) = (3*(10+-3))/(2*9)`

`x_(1,2) = (10+-3)/6 => x_1 = 13/6 ; x_2 = 7/6`

**Hence, the inequality holds for `x in (-oo,7/6)U(13/6,oo).` **

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