Standard from of the quadratic equation is anything of the form,
#y= ax^2 + bx +c#, while vertex from is in the form, #y=a(x-x_{vertex})^2 + y_{vertex}#, where #(x_{vertex},y_{vertex})# is the location of the vertex. You go from vertex form to standard form by expansion and from standard form to vertex form by "completing the square."
Starting with #y= x^2 + 10 x -5#, we first consider the "#a#" term, here #a =1#, so we are looking for a simple square of for the form
#(x+d)^2 => x^2 +2*d *x + d^2#.
Now we note that #b=10#, which is also #2 xx 5#, so we have
#y= x^2 + 2*5*x -5#.
25 would "complete the square" because #x^2 + 2*5*x + 25# is equal to the perfect square #(x+5)^2.#
We add and subtract 25 from the RHS. We are essentially adding zero to the RHS which is allowed, obviously it can't change the equation. This gives us:
#y= x^2 + 2*5*x +25 -25 -5#.
We gather the # x^2 + 2*5*x +25# into #(x+5)^2# and leave the rest out, producing: #y= (x+5)^2 -30#.
We have now completed the square, to see the vertex more clearly we can just tweak it a bit, spitting up the #+# into two #-#.
#y= (x- (-5))^2 + (-30)#. We can read the vertex off this form of the equation directly, #(-5,-30)#.
