How to solve unfactorable trinomial for x? I have to solve the trinomial $2x^2+4x+1 = 0$, which should equal $-1 \pm \frac{1}{2}\sqrt{2}$
So far, my steps are:
$$2x^2+4x=-1$$
$$2(x^2+2x)=-1$$
$$x^2+2x=\frac{-1}{2}$$
This is where I get stuck. I can take $\frac{-1}{2}$ to the left side, but  it's still unfactorable.
How can I solve this equation?
 A: Note that $(x+1)^2 = x^2+2x+1.$   I found the $1$ because it's half the middle coefficient.  
So add $1$ to both sides to get 
$$\left(x+1\right)^2 = \frac{1}{2}.$$
Enough?
A: Hint: $x^2+2x+1=\frac12$. Do you recognize the left-hand side?
A: The general procedure known as completing the square may be executed for any quadratic polynomial
$q(x) = x^2 + \alpha x + \beta \in F[x], \tag 1$
where $F$ is any field with
$\text{char}(F) \ne 2; \tag 2$
if we seek the roots of 
$q(x) = x^2 + \alpha x + \beta = 0, \tag 3$
we write this equation as
$x^2 + \alpha x = - \beta; \tag 4$
then in the light of (2) the quantity
$\left( \dfrac{\alpha}{2} \right )^2 = \dfrac{\alpha^2}{4} \in F \tag 5$
is well-defined; we may add it to each side of (4) to obtain
$\left ( x + \dfrac{\alpha}{2} \right )^2 = x^2 + \alpha x + \dfrac{\alpha^2}{4} = \dfrac{\alpha^2}{4} - \beta; \tag 6$
if now
$\exists \gamma \in F, \; \gamma^2 = \dfrac{\alpha^2}{4} - \beta, \tag 7$
then we may write
$x + \dfrac{\alpha}{2} = \pm \gamma, \tag 8$
whence
$x = -\dfrac{\alpha}{2} \pm \gamma, \tag 9$
where we note that (2) implies
$\gamma \ne -\gamma, \; \gamma \ne 0, \tag{10}$
and so the roots (9) are distinct unless $\gamma$ vanishes.  It is easy to walk these steps back and show that $x$ as in (9) satisfies (3).
In the present case we have, after dividing the quadratic equation
$2x^2 + 4x + 1 = 0 \tag{11}$
by $2$,
$x^2 + 2x + \dfrac{1}{2} = 0, \tag{12}$
we apply the above with $\alpha = 2$ and $\beta = 1/2$ and we see that
$(x + 1)^2 = x^2 + 2x + 1 = \dfrac{1}{2}, \tag{13}$
and the result
$x = -1 \pm \sqrt{\dfrac{1}{2}} = -1 \pm \dfrac{1}{2} \sqrt 2 = -1 \pm \dfrac{\sqrt 2}{2} \tag{14}$
immediately follows.
A: $$ax^2+bx+c=a\left(x^2+\frac bax+\frac ca\right)=a\left(x^2+2\frac b{2a}x+\left(\frac b{2a}\right)^2-\left(\frac b{2a}\right)^2+\frac ca\right)=a\left(\left(x-\frac b{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right).$$
The roots are such that
$$x-\frac b{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}},$$ if the quantity under the radical is non-negative.
