The equation of a parabola is (y−3)2=20(x+1) . What is the equation of the directrix of the parabola? Enter your answer in the box.THE EQUATION PLEASE. NOT the answer.

[tex]\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ 4p(y- k)=(x- h)^2 \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ (y-3)^2=20(x+1)\implies [y-\stackrel{k}{3}]^2=4(\stackrel{p}{5})[x-(\stackrel{h}{-1})]~\hfill \begin{cases} \stackrel{vertex}{(-1,3)}\\ p=5 \end{cases}[/tex]now, is noteworthy that, this parabola has the "y" squared, namely is a horizontal parabola, with a vertex at (h,k), in this case (-1,3).So, our "p" distance is a positive number 5, when "p" is positive for a horizontal parabola, that means the parabola is opening towards the right-hand-side, whilst the directrix is in the opposite direction.so, to get the directrix, we'll have to, from the vertex (-1,3), move towards the left-hand-side "5" units over, that will land us at (-6, 3), there we'll find the vertical line of x = -6.