Question 434435
It depends on what method your
teacher was trying to show you
I'll use the method known as 
"completing the square"
{{{ y^2 - 4y - 45 = 0 }}}
Add {{{45}}} to both sides
{{{ y^2 - 4y = 45 }}}
Take {{{1/2}}} of the coefficient of {{{y}}},
square it, and add it to both sides
{{{ y^2 - 4y + (-4/2)^2 = 45 + (-4/2)^2 }}}
{{{ y^2 - 4y + 4 = 45 + 4 }}}
{{{ y^2 - 4y + 4 = 49 }}}
Each side is now a perfect square
{{{ ( y - 2 )^2 = 7^2 }}}
Take the square root of both sides (both (+) root and (-) root)
{{{ y - 2 = 7 }}}
(1) {{{ y = 9 }}}
and
{{{ y - 2 = -7 }}}
(2) {{{y = -5 }}}
-----------
You can rewrite (1) and (2) as:
(1) {{{ y - 9 = 0 }}}
(2) {{{ y + 5 = 0 }}}
Now you can factor the equation:
{{{ y^2 - 4y - 45 = (y - 9)*(y+5) }}}