We start with #4x^2-17x-16=y#

#4x^2-17x-16# cannot be factored, so we will have to complete the square. To do that, we first have to make the coefficient of #x^2# #1#. That makes the equation now #4(x^2-17/4x-4)#.

The way completing the square works is, because #x^2-17/4x# isn't factorable, we find a value that *makes* it factorable. We do that by taking the middle value, #-17/4x#, dividing it by two and then squaring the answer. In this case it would look this: #(-17/4)/2#, which equals #-17/8#. If we square it, that becomes #289/64#.

We can rewrite the equation as #4(x^2-17/4x+289/64-4)#, but we can't just stick a number into an equation and not add it on both sides. We could add #289/64# to both sides, but I would prefer to just add #289/64# and then immedietly subtract it.

Now, we can rewrite this equation as #4(x^2-17/4x+289/64-289/64-4)#. Because #x^2-17/4x+289/64# is factorable, I can rewrite it as #(x-17/8)^2#. Putting it together we have #4(x-17/8)^2-289/64-4# or #4(x-17/8)^2-545/64#. The last step is to multiply #-545/64# by #4#.

The final form is #4(x-17/8)^2-545/16#