given $4x^2−4x-5=0$
we all know the solution
but what my teacher showed me is different
after we get the
\begin{align*} x & = \frac{4 \pm \sqrt{96}}{8}\\ x & = \frac{4 \pm \sqrt{4 \cdot 24}}{8}\\ x & = \frac{4 \pm 2\sqrt{4 \cdot 6}}{8}\\ x & = \frac{4 \pm 4\sqrt{6}}{8}\\ x & = \frac{4(1± \sqrt{6})}{8}\\ \end{align*} he canceled $4$ and $8$ so the $4$ becomes $1$ and the $8$ becomes $2$
$$x = \frac{1 \pm \sqrt{6}}{2}$$
Explain to me how did he do that?
\begin{align*}\frac{4 \pm \sqrt{96}}{8} &= \frac{4 \pm \sqrt{16\cdot6}}{8} \\ &=\frac{4 \pm \sqrt{16}\sqrt{6}}{8} \\ &=\frac{4 \pm 4\sqrt{6}}{8} \\ &=\frac{1 \pm \sqrt{6}}{2} \end{align*}
By the way, the correct solutions of $3x^2-2x+1$ would be:
$$\frac{2\pm\sqrt{4-12}}{6}$$
but these solutions are not real numbers.