Q.Solve $x^2 - 6x + 7 = 0$ by the method of perfect squares (completing the square).

This does not factorize with small integers, so completing the square is the right approach here.

$x^2-6x=-7 \Rightarrow x^2-6x+9=-7+9=2 \Rightarrow (x-3)^2=2 \Rightarrow x=3\pm\sqrt2$.

Since $a=1$ already, no division is needed. Move the constant to the right: $x^2-6x=-7$.

Half the coefficient of $x$ is $\dfrac{-6}{2}=-3$; its square is $(-3)^2=9$. Add $9$ to both sides: $x^2-6x+9=-7+9=2$.

The left side is now a perfect square: $(x-3)^2=2$.

Take the square root of both sides (with $\pm$): $x-3=\pm\sqrt2$, so $x=3\pm\sqrt2$.

Check via the discriminant route: $D=(-6)^2-4(1)(7)=36-28=8$, and $\sqrt{8}=2\sqrt2$, so Sridharacharya's formula gives $x=\dfrac{6\pm2\sqrt2}{2}=3\pm\sqrt2$ — identical to the perfect-square result.

Alternative — Sridharacharya's formula directly: $a=1,b=-6,c=7$, $D=36-28=8$, $x=\dfrac{6\pm\sqrt8}{2}=\dfrac{6\pm2\sqrt2}{2}=3\pm\sqrt2$ — same two roots reached without explicitly completing the square.

A very common error is adding $\left(\frac{b}{2}\right)^2$ using the ORIGINAL $b$ from the equation without first normalizing $a$ to $1$ when $a \neq 1$ — this only works directly when $a=1$, as here; for $a\neq1$ the equation must first be divided through by $a$. Another slip is forgetting the $\pm$ when taking the square root, which silently drops one of the two valid roots.