Question 15681
First of all, you can use the subsitution, but don't LOSE the equation.  What you probably mean is this:
{{{9t^2 -25t + 16 = 0}}}


This factors into:
{{{(9t-16)*(t-1) = 0 }}}


Notice that the outer times outer term is -9t, and the inner times inner is -16t, and this adds up to -25t, which is the middle term.

{{{(9t-16)*(t-1) = 0 }}}
{{{9t - 16= 0}}}
{{{9t = 16}}}
{{{t= 16/9}}}


Also, {{{t=1}}}.


Now you have to go back and solve for x, where you made the substitution {{{x^2 = t}}}

{{{x^2 = t}}}
{{{x^2 = 16/9}}}
{{{x= 4/3}}} or {{{x= -4/3}}}


{{{x^2 = t}}}
{{{x^2 = 1}}}
{{{x= 1}}} or {{{x= -1}}}


R^2 at SCC