Question 548218
We know if the quadratic function has zeros 6 and -8 then 
x-6 and x-(-8)=x+8 are both factors of the quadratic function  This means that 

{{{y=(x-6)(x+8)}}}   

Now... standard form is also known as vertex form  and is written as 
{{{y=a(x-h)^2+k}}}


We can expand our quadratic from above and then put it into this form:

{{{y=(x-6)(x+8)}}} 
{{{y=x^2+8x-6x-48}}}
{{{y=x^2+2x-48}}}

Now we must complete the square to get it into standard form.  To do this, we must add half of the b value squared ((half of 2 is 1 and 1 squared is 1 so we add one).  If we add 1, we must subtract it as well (so we are essentially not changing the equation).   
{{{y=x^2+2x-48}}}
{{{y=x^2+2x+1-48-1}}}
Now taking the first three terms, they can be factored as so:
{{{y=(x+1)^2-48-1}}}
{{{y=(x+1)^2-49}}}

there we have it:) 

Let me know if you are unsure about anything and I will make sure it is clear for you :)