You do realise that the application of the quadratic formula gives you the roots of the quadratic, ie the points on the x-axis where there parabola crosses it? That is not needed here. Clearly
x2+3 will never cross the x-axis as it is a translation vertically upwards from the basic parabola
x2 which barely touches the x-axis.
You dont need to find the roots which are evidently not real. You simply need to know from what
(x,y) point the curve begins, and you find that by plugging in
x=1 as that is the minimum value your parabola can take within
f(x) and then proceed to sketch the rest of it from that point for
xβ₯1