Key idea. Treat the repeated expression as a single unknown.

1. Put \(u=x^2-3x\). The equation becomes

   \[u^2-2u-8=(u-4)(u+2)=0,\]

   so \(u=4\) or \(u=-2\).

2. If \(u=4\), then \(x^2-3x-4=0\), or \((x-4)(x+1)=0\). Thus \(x=4\) or \(x=-1\).

3. If \(u=-2\), then \(x^2-3x+2=0\), or \((x-1)(x-2)=0\). Thus \(x=1\) or \(x=2\).

4. Each branch came from an equivalent factorisation and both branches were solved completely. All four listed values therefore satisfy the original equation, and there are no others.

Remember. A substitution is complete only after every possible value of the new variable has been converted back.