completing the square to solve equation Is it possible to use the method of completing the square to solve the equation $2x^2+18x+21=0$ ?
I have problem with how to remove the negative sign on the right side.
 A: Hint: $2x^2 + 18x + 21 = 2(x^2 + 9x) + 21 = 2\left(x+\dfrac{9}{2}\right)^2 - \dfrac{81}{2} + 21$.
A: First make the leading coefficient $1$:
$x^2 + 9x + \frac{21}{2} = 0$.
Then add and subtract $(\frac{9}{2})^2$:
$x^2 + 9x + \frac{21}{2} + (\frac{9}{2})^2 - (\frac{9}{2})^2 = 0$
Rearrange,
$x^2 + 9x + (\frac{9}{2})^2  = (\frac{9}{2})^2 - \frac{21}{2}$
Observe the LHS is a square:
$(x + \frac 92)^2 = \frac{81}{4} - \frac{42}{4} = \frac{39}{4}$
Take square root of both sides,
$x + \frac 92 = \pm \frac 12 \sqrt {39}$
Final solution:
$x = - \frac 92 \pm \frac 12 \sqrt {39}$
A: Make the leading coefficient a square avoiding fractions by multiplying $2x^2+18x=-21$ by $2$, you'll get $4x^2+36x=-42$.  Now add $81$ to complete the square:
$$4x^2+36x+81=39\iff(2x+9)^2=39\iff2x+9=\pm\sqrt{39}\iff x=\frac{-9\pm\sqrt39}{2}.$$
A: $$
2\left[\left(x+\frac92\right)^2+\frac{21}{2}-\frac{81}{4}\right]=
2\left[\left(x+\frac92\right)^2-\frac{39}{4}\right]
$$
and so on.
A: Here is a graphical solution that starts with completing the square and finishes off with a difference of squares. I hope it downloads okay for everyone.

