We are given the system of equations:

$x^2 + y^2 = 100$ $xy = 14$

To solve for $x$ and $y$, we can use these equations to find a relation between $x$ and $y$.

Step 1: Use the identity $(x + y)^2 = x^2 + y^2 + 2xy$

From the first equation, we know that $x^2 + y^2 = 100$, and from the second equation, we know that $xy = 14$. So we can substitute these into the identity:

$(x + y)^2 = x^2 + y^2 + 2xy$ $(x + y)^2 = 100 + 2(14)$ $(x + y)^2 = 100 + 28$ $(x + y)^2 = 128$

Step 2: Solve for $x + y$

Taking the square root of both sides:

$x + y = \pm \sqrt{128}$ $x + y = \pm 8\sqrt{2}$

So, $x + y$ could be either $8\sqrt{2}$ or $-8\sqrt{2}$.

Step 3: Use the quadratic equation to find $x$ and $y$

Now, let’s express $x$ and $y$ as the roots of the quadratic equation:

$t^2 - (x + y)t + xy = 0$

Substitute the values we know:

$t^2 - (\pm 8\sqrt{2})t + 14 = 0$

This is a quadratic equation in $t$ that we can solve for the values of $x$ and $y$.

Would you like me to go ahead and solve this quadratic equation for $x$ and $y$?