problem: What is the distance between the two intersections of $y=x^2$ and $x+y=1$?
solution: To find the $x$-coordinates of the intersections, substitute $x^2$ for $y$ in $x+y=1$ and solve for $x$, resulting in  \begin{align*}
  x+x^2&=1 \\
  \Rightarrow \qquad x^2+x-1&=0 \\
  \Rightarrow \qquad x&=\frac{-1\pm\sqrt{1+4}}2=\frac{-1\pm\sqrt5}2\\
  \end{align*}Using each of these coordinates to solve for $y$ gives us the intersections at $\left(\frac{-1+\sqrt5}2,\frac{3-\sqrt5}2\right)$ and $\left(\frac{-1-\sqrt5}2,\frac{3+\sqrt5}2\right)$.  Using the distance formula, we have \begin{align*}
  &\sqrt{ \left(\frac{-1+\sqrt5}{2}-\frac{-1-\sqrt5}{2}\right)^2 + \left(\frac{3-\sqrt5}2-\frac{3+\sqrt5}2\right)^2 }\\
  &\qquad=\sqrt{\left(\frac{2\sqrt5}2\right)^2 + \left(-\frac{2\sqrt5}2\right)^2}\\
  &\qquad=\sqrt{ 2\sqrt5^2 }\\
  &\qquad=\boxed{\sqrt{10}}.
  \end{align*}