Question 967930
Treat the equation through substitution this way:


{{{w=x^2}}}
-
{{{2w^2-5w-12=0}}}
{{{2w^2-5w=12}}}
{{{2(w^2-(5/2)w)=12}}}
{{{2(w^2-(5/2)w+(5/4)^2)=12+2*(5/4)^2}}}, careful about the details in this step.  Accounting is done for the factor of 2 on the left member, and adjusted for this on the right member.


{{{2(w-(5/4))^2=12+2*25/16}}}
{{{2(w-(5/4))^2=12+25/8}}}
{{{(w-(5/4))^2=6+25/16}}}
{{{(w-(5/4))^2=(96+25)/16}}}
{{{(w-(5/4))^2=121/16}}}
Square ROOT of both sides:
{{{(w-(5/4))=11/4}}}-------actually, PLUS or MINUS.
Substitute BACK for w.
{{{x^2-5/4=0+- 11/4}}}
{{{x^2=5/4+- 11/4}}}, and your other variable for the form is p=0; so you can say
{{{(x-0)^2=5/4+- 11/4}}}
OR
{{{highlight((x-0)^2=(5+- 11)/4)}}}