Question 836763
How about {{{u=sqrt(x)}}}, then {{{u^2=x}}},
{{{2u^2-11u-6=0}}}
{{{(2u+1)(u-6)=0}}}
Two solutions:
{{{2u+1=0}}}
{{{2u=-1}}}
{{{u=-1/2}}}
{{{sqrt(x)=-1/2}}}
This is not a solution.
{{{u-6=0}}}
{{{u=6}}}
{{{sqrt(x)=6}}}
{{{x=36}}}