The sum and difference of two numbers are 19 and 7, respectively. What is the sum of their squares?

The correct option is 205.

Let the two numbers be $x$ and $y$.

According to the problem, we are given two equations:
1) The sum of the two numbers is 19:
$x+y=19$
2) The difference of the two numbers is 7:
$x-y=7$

We want to find the sum of their squares, which is:
$x^2+y^2$

To find the values of $x$ and $y$, we can add the two equations together:
$(x+y)+(x-y)=19+7$
$2x=26$
$x=13$

Now, we can substitute $x=13$ back into the first equation to find $y$:
$13+y=19$
$y=19-13$
$y=6$

Now that we have the values of both numbers, $x=13$ and $y=6$, we can calculate the sum of their squares:
$x^2+y^2={13}^2+6^2$
$x^2+y^2=169+36$
$x^2+y^2=205$

Alternatively, we can use the algebraic identity:
$(x+y{)}^2+(x-y{)}^2=2(x^2+y^2)$

Substituting the given values into the identity:
${19}^2+7^2=2(x^2+y^2)$
$361+49=2(x^2+y^2)$
$410=2(x^2+y^2)$
$x^2+y^2=205$