Have two things bugging me -

F(x)=(x-3)^2-5

Wanted to find the x intercepts using the quadratic formula

I wind up 6+- sqrt 20, how can I find x intercepts from that.

I found the correct x intercepts using the 0=x^2-6x+4

Then take the sqrt of 4 =2

then (x-3)=2 (x-3) = -2

x=5 x=1

That matches the trace in my calculator, so I am sure this is correct but I just wish I could have found them using the quadratic method.

The other thing that seems weird is the domain and range.

my calculator shows all real numbers for domain, but the lowest number on the range was (-5, oo). Would like to see another way to prove the range.