$\displaystyle 2x^2 -7x +3 = 0$

$\displaystyle 2x^2 -7x = -3$

$\displaystyle \frac{\2x^2}{2} -\frac{\7}{2}x = -\frac{\3}{2}$

Take one half of -$\displaystyle \frac{\7}{2}x$ and then square that answer which = $\displaystyle \frac{\49}{16}$ and add to both sides

$\displaystyle x^2 -\frac{\7}{2}$ +$\displaystyle \frac{\49}{16}$= -$\displaystyle \frac{\3}{2}$ + $\displaystyle \frac{\49}{16}$

Combine like terms on the right to get

$\displaystyle \frac{\25}{16}$

and then subtract that from both sides to get

$\displaystyle x^2 -\frac{\7}{2} + \frac{\24}{16}=0$

$\displaystyle \frac{\24}{16}$ can be reduced to $\displaystyle \frac{\3}{2}$ so that

$\displaystyle x^2 -\frac{\7}{2} + \frac{\3}{2}=0$

Eliminate the fractions by multiplying it all by 2 and then instead of completing a square I've completed a large useless circle as I arrive back at the original equation.

$\displaystyle 2x^2 -7x +3=0$

Now, I also used $\displaystyle \frac{\49}{4}$ as the square of half of $\displaystyle -\frac{\7}{2}$ but eventually it also ends up as

$\displaystyle \frac{\3}{2}$ on the left hand side of the equation.

I did this problem with the quadratic formula in about 45 seconds but I have to learn it this way. If nothing else my LaTex skills have been markedly improved in this post.

Any help is appreciated. Thanks.