How to solve Y^2-8y-16

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