How to systematically find roots of $x^2-x-132 = 0$ with Po-Shen Loh's method?

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Po-Shen Loh in his famous video shows how to systematically find quadratic equation's roots.

He find the roots for following quadratic equation.

$x^2-8x+12 = 0$

Product: 12, Sum: 8

He divides sum by half i.e. 8/2 = 4

Proceeds to find roots: $4 - u$ & $4 + u.$

$ (4 - u) * (4 + u) = 12$

$16 - u^2 = 12 $

$16 - 12 = u^2$

$ 4 = u^2$

$u = \pm2$

Now, if we substitute value of $u$ in $4 - u$ & $4 + u$, we get, $2$ and $6$

But, this methods doesn't yield result for $x^2-x-132 = 0$

Let's divide sum by half i.e. $(1) * 1/2 = 1/2$

Proceeds to find roots: $1/2-u$ & $1/2+ u$

$ (1/2 - u) * (1/2 + u) = - 132$

$1/4 - u^2 = - 132 $

$1/4 + 132 = u^2$

$ 529/4 = u^2$

$ 132.25 = u^2$

$u = ?$

How to systematically find roots of $x^2-x-132 = 0$ with above method?

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$$u^2=132.25$$

$$u=\pm11.5$$

$$\frac{1}{2}+11.5=12$$

$$\frac{1}{2}-11.5=-11$$

$$(x-12)(x+11)=0$$

The $-132$ does change things, by the way.