Question 1095268
ONE WAY:
Maybe we know about solving quadratic equations.
Sometimes we can solve a quadratic equation such as {{{z^2+bz+c=0}}} by factoring if we find two integers {{{p}}} {{{q}}} such that {{{p+q=-b}}} and {{{pq=c}}} .
Then those values for {{{p}}} and {{{q}}} are the solutions of the equation and the equation is really
{{{z^2-(p+q)z+pq=(z-p)(z-q)=0}}} .
The solutions to a quadratic equation of the form {{{z^2+bz+c=0}}} are always two numbers whose sum and product can be found are {{{-b}}} and {{{c}}} respectively.
Unfortunately, sometimes those numbers are irrational, or even imaginary, and then factoring is not an option.
Then, w must use algebra to "complete the square" and then solve, or apply the dreaded quadratic formula
that says the solutions to an equation of the form {{{ax^2+bx+c=0}}} are given by
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
To avoid confusion, I used {{{z}}} for the variable instead of {{{x}}} , and I prefer equations where the leading coefficient is {{{a=1}}} , so I wrote my equation as {{{z^2+bz+c=0}}} .
I will make {{{b=-(p+q)=-66}}} and {{{c=pq=9}}} to get {{{z^2-66z+9=0}}}. 
The solutions will be the numbers {{{p}}} and {{{q}}} (or {{{x}}} and {{{y}}} ) that add up to {{{x+y=66}}} and whose product is {{{x*y=9}}} .
The quadratic formula tells me that the solution are given by
{{{z = (-(-66) +- sqrt((-66)^2-4*1*9 ))/(2*1)=(66 +- sqrt(4356-36))/2=(66 +- sqrt(4320))/2=(66 +- sqrt(144*30))/2=(66 +- 12sqrt(30))/2=33 +- 6sqrt(30)}}}
The solutions to the equation are {{{(33+6sqrt(30))}}} and {{{(33-12sqrt(30))}}}
One is {{{x}}} and the other is {{{y}}} , but there is no way to guess which was intended to be {{{x}}} and which was intended to be {{{y}}} .
{{{x^2}}} and {{{y^2}}} are
{{{(33+6sqrt(30))^2=33^2+2*33*6sqrt(30)+36*30=1089+396sqrt(30)+1080}}} and
{{{(33-6sqrt(30))^2=33^2-2*33*6sqrt(30)+36*30=1089-396sqrt(30)+1080}}}
The possible answers are
{{{x^2-y^2=1089+396sqrt(30)+1080-(1089-396sqrt(30)+1080)
=1089-1089+396sqrt(30)+396sqrt(30)+1080-1080=highlight(792sqrt(30))}}} and {{{x^2-y^2=1089-396sqrt(30)+1080-(1089+396sqrt(30)+1080)
=1089-1089-396sqrt(30)-396sqrt(30)+1080-1080=highlight(-792sqrt(30))}}}
 
ANOTHER APPROACH:
We know that {{{x^2-y^2=(x+y)(x-y)}}} . If we only knew the value of {{{(x-y)}}} we could easily find the value of {{{x^2-y^2}}} 
 
We know that {{{(x-y)^2=x^2+y^2-2xy}}} , but to calculate {{{(x-y)^2}}} we would need the value of {{{x^2+y^2}}} 
 
We know that {{{(x+y)^2=x^2+y^2+2xy}}} and we know the values of {{{(x+y)=66=2*3*11}}} and {{{xy=9=3^2}}}
Substituting the known values we get {{{66^2=x^2+y^2+2*9}}} --> {{{66^2=x^2+y^2+18}}} --> {{{x^2+y^2=66^2-18=4338}}}

Now we can use that value of to calculate the values of {{{(x-y)^2}}} and {{{(x-y)}}} , and from that find the value of {{{x^2-y^2}}}
 
{{{(x-y)^2=x^2+y^2-2xy=4338-2*9=4338-18=4320=30*144=30*12^2}}}
 
Then {{{(x-y)=" " +- sqrt(30-12^2)=" " +- 12sqrt(30)}}}
 
Multiplying times {{{(x+y)= 66*(" " +- 12sqrt(30)) =highlight(" " +- 792sqrt(30))}}}