Finding a real solution to a quadratic I am suppose to find all the solutions to this problem, I think some theorem states that there can only be as many solutions to the problem as the highest degree. I know that calculus reinforces this so I know that
$2x^2 + 4x + 1 = 0$
Can have at most two solutions. In calculus this is proven by the derivative being zero at only somewhere. I can't remember and it isn't important yet.
Anyways I have no idea what to do with this problem. I don't think I can factor it conventionally because of the 2 coefficient so what is the method at this point? I tried guessing and it didn't work at all for -2 - 3.
 A: $$ax^2 + bx + c = 0, \text{ roots?}$$

This is where the quadratic formula comes in handy (you should memorize this!):

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The discriminant d of the quadratic formula is the term $d= (b^2 - 4ac)$. When $b^2\geq 4ac$, you have two real roots. 
At the very least, try to memorize how to compute the discriminant. That will allow you to easily determine whether or not a quadratic equation has real roots.
A: If you don't want to use the formula, you can do the following.(Actually this is how the formula is derived though. I used to use this method when I didn't memorize the formula.)
$2x^2+4x+1=0$
$x^2+2x+1/2=0$
$(x+1)^2-1/2=0$
$(x+1)^2=1/2$
$x+1=+1/\sqrt 2$ or $-1/\sqrt 2$
A: Use the Quadratic Formula. One cannot expect all quadratic polynomials to factor "nicely." The Quadratic Formula needs to become a completely standard tool in your arsenal. 
A: The quadratic formula works for all quadratic equations. For any equation [in the form] $ax^2 + bx + c = 0$, this is true:$$x = {-b \pm \sqrt{b^2 - 4ac} \over 2a}$$
A: You should definitely commit the formula to memory. Here's a cool video to the tune of row-row-row-your-boat, but there are plenty more out there. http://www.youtube.com/watch?v=HRcj9slciqM. Once you get on youtube, search around as there are lots of great videos in there that show you how to use the formula.
