Finding the inverse of a quadratic function?

[SoConfused__]

Hello, I am sorry for removing the template but it is not an actual problem that I need help with. Well it is, but it should be pretty quick and straight forward.

The problem is located here: http://www.cdli.ca/courses/math3103/unit05_org01_ilo04/images/3-less9.gif

(It's a bit difficult to type out.)

I'm having difficulty with only 1 step that I need straightened out.

On the second last line, where did the 1/2 outside of the square root function come from? And Why does it appear there? I am thinking the 1/2 just got moved out of it, because it was under x+3.

Please help, thank you so much.

 

[Dick]

They wrote (x+3)/-2=(x+3)*2/((-2)*2)=(2x+6)/(-4)=(1/4)*((2x+6)/(-1))=(1/4)*(-2x-6). Then they took sqrt(1/4)=1/2 and moved that out.

 

[snshusat161]

Multiply both numerator and denominator by -2 inside the square root

y + 1 = \pm \sqrt{\frac{-2(x + 3)}{- 2 (-2)}

 

[snshusat161]

Multiply both numerator and denominator by -2 inside the square root

y + 1 = \pm \sqrt{\frac{-2x-6}{4}