Yes that is what I was doing Chris, I just didn't halve them because I wanted to do it in my head. :))
Here is another way Sweaty, although it is probably over your head.
You can use calculus and get a lot of accuracy very quickly using
Newton's method of approximating roots.
I derive this formula every time I use it for a have a really bad memory but most people just memorise the formula.
$$\boxed{x_2=x_1-\frac{f(x_1)}{f'(x_1)}}$$
$$\\let\;\;x=\sqrt{6}\\
x^2=6\\
x^2-6=0\\
$ the idea is to let $ f(x)=x^2-6 $ and solve for $y=0\\\\
f(x)=x^2-6\\
f'(x)=2x\qquad $ that is calculus $\\
$now we know that $\;\; 2 $So my first estimate will be $x_1=\textcolor[rgb]{1,0,0}{\mathbf{2.5}}\\\\
x_2=2.5-\frac{2.5^2-6}{2*2.5}\\\\
x_2=\textcolor[rgb]{1,0,0}{\mathbf{2.45}}$$
$$\\\mbox{second use of the formula}\\\\
x_3=2.45-\frac{2.45^2-6}{2*2.45}\\\\
x_3=\textcolor[rgb]{1,0,0}{\mathbf{2.449489796}}\\\\
$third use of the formula$\\\\
x_4=2.449489796-\frac{2.449489796^2-6}{2*2.449489796}\\\\
x_4=\textcolor[rgb]{1,0,0}{\mathbf{2.449510072}}\\\\
$The value is already correct to 4 decimal places }
\sqrt{6}=2.4495\\\\\\
$With each iteration the accuracy will increase.$$$
You will learn and understand this when you do calculus - something to look forward too :))
