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. 
$$
d(x) = \sqrt{(x+1)^2 + \big(-(x+1)^2 + 2\big)^2}
$$

Let $(x+1)^2=t$, where $t\geq0$.
Thus, we need to solve $t+(t-2)^2=14$.
I hope now it's clear. 
