Question 343922
Use a substitution, {{{u=x^2-2x}}}
{{{(x^2 -2x)^2 -11(x^2 -2x) +24=0}}}
{{{u^2 -11u+24=0}}}
{{{(u-8)(u-3)=0}}}
Two solutions in {{{u}}}:
{{{u-8=0}}}
{{{u=8}}}
{{{x^2-2x=8}}}
{{{x^2-2x-8=0}}}
{{{(x-4)(x+2)=0}}}
Two solutions in {{{x}}}:
{{{x-4=0}}} 
{{{highlight(x=4)}}}
.
.
{{{x+2=0}}}
{{{highlight(x=-2)}}}
.
.
.
{{{u-3=0}}}
{{{u=3}}}
{{{x^2-2x=3}}}
{{{x^2-2x-3=0}}}
{{{(x-3)(x+1)=0}}}
Two solutions in {{{x}}}:
{{{x-3=0}}}
{{{highlight_green(x=3)}}}
.
.
.
{{{x+1=0}}}
{{{highlight_green(x=-1)}}}
.
.
.
Four solutions: (-2,-1,3,4)