complete the square

I think this is an odd question.

Usually completing the square is done as a quadratic equation solving technique.  

But if you just want to know what constant needs to be added to make it a perfect square then this is what I'd do.

(it is very similar to what heureka has done)

$$4x^2-4x-27\\\\
=4(x^2-x-\frac{27}{4})\\\\
=4(x^2-1x+\frac{1}{4}-\frac{28}{4})\\\\
=4(x^2-1x+\frac{1}{4}-\frac{28}{4})\\\\
=4(x^2-1x+\frac{1}{4})-28\\\\
=2^2(x^2-\frac{1}{2}x+\frac{1}{4})-28\\\\
=2^2(x-\frac{1}{2})^2-28\\\\
=\left(2(x-\frac{1}{2})\right)^2-28\\\\
=\left(2x-1\right)^2-28\\\\$$

So the expression would be a perfect square if 28 were added.

$$\\4x^2-4x-27 \\
=4(x^2-x)-27\\
=4\left(x^2-x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right) -27\\
=4\left( \left( x-\frac{1}{2}\right)^2-\frac{1}{4}\right) -27\\
=4\left(x-\frac{1}{2}\right)^2 - \frac{4}{4}-27\\
=4\left(x-\frac{1}{2}\right)^2 - 1-27\\
=4\left(x-\frac{1}{2}\right)^2 -28\\$$

.