Question 757031
x = (y^2) + -9 and x - 4y - 12 = 0 

{{{x=4y+12}}} from second equation. Substitute into the first equation.

{{{4y+12=y^2-9}}}
{{{y^2-4y-9-12=0}}}
{{{y^2-4y-21=0}}}
You should no longer be stuck.  The lefthand side is factorable.





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Just a little bit of review:

{{{(x+h)(x+k)=x^2+(h+k)x+hk}}}


If you find some square trinomial as any general {{{x^2+ax+b}}} and you want to factor it, try thinking of it as {{{x^2+ax+b=x^2+(h+k)x+hk}}}
So you are looking for two numbers, h and k, so that hk=b and h+k=a.  
What you do is look for all the two-number factorizations for b, and find which ones give you a sum of a.  We then test each combination until we find something that works.  There is a somewhat more analytical way also, but the search and test method is usually enough.