you can't solve it because it isn't equal anything
but we can factor
since the quadratic coefient (number in front of y^2 term) is 1
we can do this
for ax^2+bx+c
when a=1
what 2 numbers multiply to c and add to b
what 2 numbers multiply to -16 and add to -8
no numbers
if you had y^2-8y+16 then that would be (x-4)(x-4)
but no
ok, so we need to complete the squaer
so
y^2-8y-16
take 1/2 of linear coefient and square it
-8/2=-4, -4^1=16
add negative and poisitve to it
y^2-8y+16-16-16
factor
(y-4)^2-16-16
(y-4)^2-32
then we can force factor again
remember difference of 2 perfect square
a^2-b^2=(a+b)(a-b)
√32=4√2
so
(y-4)^2-(4√2)^2=(y-4+4√2)(y-4-4√2)
the factore form would be (y-4+4√2)(y-4-4√2)
not equal to anything tho so we can't solve
