Start by factoring: $$ \begin{aligned} |6x^2-47x+15-28x| &= |6x^2-75x+15| \\ &= |3(2x-5)(x-1)| \\ \end{aligned} $$ For the expression to be prime, the value of $|3(2x-5)(x-1)|$ must be 1 (since prime numbers by definition have no factors other than 1 and the number itself) which is achieved when the expression inside the absolute value is either 1 or -1.
Case 1: $2x-5=1$ and $x-1=1$. Solving, we get $x = \frac{6}{2} = 3$, but this does not result in the expression being prime.
Case 2: $2x-5=-1$ and $x-1=-1$. Solving, we get $x = \frac{4}{2} = 2$, which does not result in a prime expression.
Case 3: $2x-5=1$ and $x-1=-1$. Solving, we get $x = \frac{4}{2} = 2$, which does not result in a prime expression.
Case 4: $2x-5=-1$ and $x-1=1$. Solving, we get $x = \frac{4}{2} = 2$, which does not result in a prime expression.
Case 5: $2x-5=-1$ and $x-1=-1$. Solving, we get $x = \frac{4}{2} = 2$, which does not result in a prime expression.
The only remaining case is $3(2x-5)(x-1) = -1$. Solving, we get $x=\boxed{3}$, which does result in a prime expression ($|-33| = 33$).
What is the greatest integer $x$ such that $|6x^2-47x+15-28x|$ is prime?
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