Question 383987
1.{{{y=x^2}}}
2.{{{y=x+12}}}
Substitute eq. 1 into eq. 2,
{{{x^2=x+12}}}
{{{x^2-x-12=0}}}
{{{(x-4)(x+3)=0}}}
Two solutions:
{{{x-4=0}}}
{{{x=4}}}
Then 
{{{y=4^2}}}
{{{y=16}}}
.
.
.
{{{x+3=0}}}
{{{x=-3}}}
Then,
{{{y=(-3)^2}}}
{{{y=9}}}
.
.
.
{{{drawing(300,300,-5,5,-5,20,grid(1),circle(-3,9,0.2),circle(4,16,0.2),graph(300,300,-5,5,-5,20,0,x^2,x+12))}}}