Answer:
Step-by-step explanation:
[tex]y=4+x[/tex] ............................................ (1)
[tex]y=x^2+x-12[/tex] ............................................ (2)
now equate both (1) and (2)
[tex]4+x=x^2+x-12[/tex]
[tex]=>x^2=16[/tex]
[tex]x=-4,4[/tex]
now substitute the values of x ,
when [tex]x=-4[/tex] , y becomes 0
when [tex]x=4[/tex] , y becomes 8
so, solution set:
[tex](-4,0)[/tex] and [tex](4,8)[/tex]