Solve for X: (2x+1)^2=16

$$\\(2x+1)^2=16\quad|\quad\sqrt{}\\
2x+1=4\quad|\quad-1\\
2x=3\quad|\quad:2\\
x=\frac{3}{2} \quad or \quad x=1.5$$

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Don't forget the other solution!

If (2x+1)² = 16 then 2x+1 = ±4

Taking the negative value we have 2x+1 = -4;  2x = -5;  x = -5/2 or x = -2.5

I think that you have forgotten one of the answers heureka.

$$(2x+1)^2=16\\

2x+1=4\quad or \quad 2x+1=-4\\

2x=3\quad \:\:\:\:or \quad 2x=-5\\$$

 

$$x=1\frac{1}{2}\quad or \quad x=-2\frac{1}{2}\\$$

I was your looking at your LaTex code heureka,  I've not seen the \quad command before.

I used it in my answer.  Thank you.