Question 400380
Ok, I am working on graphing the quadratic 5x^2+10x+2=0 but I do not full understand how to do it

You should first note that this is a parabola which opens upward because the coefficient of the x^2 is positive.
In order to graph it, you need to put it in this form: y=(x-h)^2+k, to find its vertex and x & y intercepts.For problems like this it would be very helpful to use a graphing calculator to see what the curve looks like.

y=5x^2+10x+2
completing the square
y=5(x^2+2x+1)-5+2
y=5(x+1)^2-3
This gives the (x,y) coordinates of the vertex (-1,-3)
To find the y-intercept, set x=0,  then solve for y which by inspection you can see it is=2
To find the x-intercepts, set y-0, then solve for x
(x+1)^2=3/5
(x+1)=sqrt(3/5)
x=-1+-sqrt(3/5)
x=-1.77and x=-.22

You now have coordinates of the vertex, the y-intercept and the x intercepts to help you draw the curve.
It should look similar to the graph below:

{{{ graph( 300, 200, -6, 5, -10, 10, 5x^2+10x+2) }}}