Question 992750
<pre>
Or you can use u-substitution:

{{{2(x/(x+2))^2 - 3((x)/(x+2)) - 9 = 0}}}

Let {{{u=x/(x+2)}}}

then the above becomes:

{{{2u^2 - 3u - 9 = 0}}}

{{{(u-3)(2u+3)=0}}}

{{{u-3=0}}};  {{{2u+3=0}}}

{{{u=3}}};  {{{2u=-3}}}

             {{{u=-3/2}}}

Then since  {{{u=x/(x+2)}}}

{{{x/(x+2)=3}}};   {{{x/(x+2)=-3/2}}}

Mutiply both sides by x+2

{{{x=3(x+2)}}};   {{{2x=-3(x+2)}}}

{{{x=3x+6}}};      {{{2x=-3x-6}}}

{{{-2x=6}}};        {{{5x=-6}}}

{{{x=-3}}};          {{{x=-6/5}}}

Edwin</pre>