Question 753095
solve y^2-x+2y+3=0 
complete the square:
(y^2+2y+1)-1-x+3=0
(y+1)^2=x-2
This is an equation of a parabola that opens rightward with vertex at (2,-1)
Its basic equation: (y-k)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex
see graph below:


{{{ graph( 300, 300, -6, 10, -10, 10, (x-2)^.5-1,-(x-2)^.5-1) }}}