Find the coordinates of the points on $y=-(x+1)^2+4$ that have a distance of $\sqrt {14}$ to $(-1,2)$ Create a function that gives the distance between the point $(-1,2)$  the graph of $$y=-(x+1)^2+4.$$ Find the coordinates of the points on the curve that have a distance of $\sqrt {14}$ units from the point $(-1,2)$.  
I know that the $x$-intercepts are $x=1$ and $x=-3$, and that the vertex is $(-1,4)$. I'm trying to use the distance formula by equating $\sqrt{14}$ to the function, but that is not getting me anywhere. 
$$
d(x) = \sqrt{(x+1)^2 + \big(-(x+1)^2 + 2\big)^2}
$$

You need to solve $$\sqrt{(x+1)^2 + \big(-(x+1)^2 + 2\big)^2}=\sqrt{14}.$$ Squaring both parts and substituting $u=(x+1)^2$ we get a quadratic equation $u^2-3u-10=0$ which has two roots, $u=5$ and $u=-2$. Only $(x+1)^2=5$ is possible for real numbers so we get $x=\sqrt5-1$ and $x=-\sqrt5-1$
