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).
