Question 621376
Sometimes the easiest way to solve a quadratic equation is by factoring, and this is a great example of that.
{{{y^2-6y+9}}} should be recognized as the square of a binomial, because
{{{y^2}}} is the square of {{{y}}},
{{{+9}}} is the square of {{{3}}} and of {{{-3}}},
and {{{-6y=2*y*(-3)}}} is double the product of {{{y}}} and {{{-3}}}.
So, {{{y^2-6y+9=(y-3)^2}}} and then
{{{y^2-6y+9=0}}} ---> {{{(y-3)^2=0}}} ---> {{{y-3=0}}} ---> {{{highlight(y=3)}}}
Not all quadratic equations can be solved by factoring, but when it is possible, it is easier than dealing with the dreaded quadratic formula.