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 $abc$.
First we simplify to $x^2-11x+6 = 0$. Now we solve for $x$ using $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (note that $a=1$, $b=-11$, and $c=6$). The roots of this equation are
\[x = \frac{11\pm \sqrt{11^2-24}}{2}=\frac{11\pm \sqrt{97}}{2}.\]
Therefore $a=11$, $b=97$, and $c=2$, so our answer is $abc=\boxed{2134}$.
To solve the given quadratic equation $x(x-3)=1+8x-5$, we first simplify it by expanding the left-hand side:
$x(x-3)=1+8x-5$
$x^2-3x=1+8x-5$
Next, we combine like terms:
$x^2-3x=8x-4$
Rearranging the equation, we have:
$x^2-3x-8x+4=0$
$x^2-11x+4=0$
Now, we can use the quadratic formula to find the roots of the equation. The quadratic formula states that for an equation of the form $ax^2+bx+c=0$, the roots can be found using the formula:
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
For our equation, $a=1$, $b=-11$, and $c=4$. Substituting these values into the quadratic formula:
$x=\frac{-(-11)\pm\sqrt{(-11)^2-4(1)(4)}}{2(1)}$
Simplifying further:
$x=\frac{11\pm\sqrt{121-16}}{2}$
$x=\frac{11\pm\sqrt{105}}{2}$
Since $105$ is not a perfect square, the roots of the equation cannot be simplified further. Therefore, the roots can be expressed as $\frac{11\pm\sqrt{105}}{2}$.
From this, we can determine that $a=11$, $b=105$, and $c=2$. Finally, we find the product of $a$, $b$, and $c$:
$abc=11\times105\times2=2310$
Therefore, $abc=2310$.