The roots of the quadratic equation $x(x-3)=1+8x-5$ may be expressed in the form $\frac{a+\sqrt{b}}{c}$ and $\frac{a-\sqrt{b}}{c}$, where $a$, $b$, and $c$ are prime numbers. Find $abc$.
\(x(x-3)=1+8x-5\)
Expand and simplify:
\(x^2-3x=1+8x-5 \\ \iff x^2-11x+4=0\)
\(\text{Now, use the quadratic formula: } \) \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(\text{Where: a=1,b=-11,c=4}\)
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