Hi

Using the vertex form of a parabola,  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



 

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