Question 415258
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Hi
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
x^2+5+2x = 0 
y = x^2+ 2x + 5    |completing Square to put into vertex form for graphing purposes
 y = (x+1)^2 - 1 + 5
 y = (x+1)^2 + 4   |Vertex is (-1,4) and line of symmetry is x = -1 and(0,5) the y-intercept
NO real solutions, Parabola does not cross the x-axis.
0 = (x+1)^2 + 4
 -1 ± 2i = x
{{{drawing(300,300,   -6, 6, -6, 6,  blue(line(-1,6,-1,-6))  , grid(1),
circle(-1, 4,0.3),
circle(0,5,0.3),
graph( 300, 300, -6, 6, -6, 6,0, x^2+5+2x ))}}}