Q.Solve $2x^2 - 4x - 3 = 0$ by the method of perfect squares.

Here $a=2 \neq 1$, so we divide through by $2$ first before completing the square.

Divide the whole equation by $a=2$: $x^2-2x-\dfrac32=0$.

Move the constant to the right side: $x^2-2x=\dfrac32$.

Half the coefficient of $x$ is $\dfrac{-2}{2}=-1$; its square is $1$. Add $1$ to both sides: $x^2-2x+1=\dfrac32+1=\dfrac52$.

Left side is a perfect square: $(x-1)^2=\dfrac52$.

Take square roots: $x-1=\pm\sqrt{\dfrac52}=\pm\dfrac{\sqrt{10}}{2}$, so $x=1\pm\dfrac{\sqrt{10}}{2}$. …

Alternative — Sridharacharya's formula on the original equation directly: $a=2,b=-4,c=-3$, $D=16+24=40$, $x=\dfrac{4\pm\sqrt{40}}{2 …

Forgetting to divide the CONSTANT term by $a$ as well when normalizing (only dividing the $x^2$ and $x$ terms) is a very common error — every term of the equation must be divided by $a$, including $c$. Also, $\sqrt{40}$ should be simplified to $2\sqrt{10}$ before dividing, otherwis …