Please help ASAP!!!!

Answer:
A. [tex]x=6[/tex] or [tex]x=1[/tex]
Step-by-step explanation:
The given function is:
[tex]y=x^2-7x+6[/tex]
When y=0; we have:
[tex]x^2-7x+6=0[/tex]
[tex]x^2-6x-x+6=0[/tex]
Factor by grouping:
[tex]x(x-6)-1(x-6)=0[/tex]
[tex](x-6)(x-1)=0[/tex]
Either [tex](x-6)=0[/tex] or [tex](x-1)=0[/tex]
Either [tex]x=6[/tex] or [tex]x=1[/tex]