Answer
The roots of the equation ${{y}^{2}}-6y-4=0$ by the use of completing the square method are:
$y=\left\{ 3+\sqrt{13}\text{ ,}3-\sqrt{13} \right\}$.
Work Step by Step
In completing the square method, convert the equation into the form of ${{\left( a+b \right)}^{2}}$ or ${{\left( a-b \right)}^{2}}$.
We have:
${{y}^{2}}-6y-4=0$
${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
But $a=y$, and
$\begin{align}
& 2ab=6y \\
& 2b=6 \\
& b=3
\end{align}$
So, ${{b}^{2}}={{3}^{2}}=9$
Now, add 9 to both sides of ${{y}^{2}}-6y-4=0$,
$\begin{align}
& {{y}^{2}}-6y-4+9=9 \\
& {{y}^{2}}-6y+9=9+4 \\
& {{y}^{2}}-6y+9=13
\end{align}$
But, the equation ${{y}^{2}}-6y+9$ is the complete square of ${{\left( y-3 \right)}^{2}}$,
${{\left( y-3 \right)}^{2}}=13$
Taking the square root on both sides,
$y-3=\pm \sqrt{13}$
Therefore, the roots are $y=3+\sqrt{13}\text{ or }y=3-\sqrt{13}$.
