There are a couple of ways to do this.
(1)
Bring the constant over to join the other terms x2+2x-3, then factorise: (x+3)(x-1)
Let y=x2+2x-3=(x+3)(x-1) and draw the graph: The x intercepts are -3 and 1. The y intercept is just the constant term -3, which is below the intercepts, so the graph is an upright parabola cutting the x axis at the intercepts and cutting the y axis at -3. We want to know when y>0, that is, when the curve lies above the x-axis. The graph shows that the positive part of the graph is when x<-3 and x>1.
(2)
We factorise as we did for the above graph: x2+2x>3, x2+2x-3>0, (x+3)(x-1)>0. This expression is positive when x+3 and x-1 are both positive or both negative, so we solve inequalities:
x+3>0 and x-1>0, which gives us x>-3 and x>1. So since 1 is greater than -3, we only need x>1.
x+3<0 and x-1<0 gives us x<-3 and x<1. So since -3 is less than 1, we only need x<-3.
We can show this on a number line. So draw a horizontal number line so that it includes -3 and 1.
On the left mark the number line (perhaps using a coloured pen) so as to highlight all points to the left of -3 and all points to the right of 1 (you could use -5 to 5 as the range of values).
It is normal to mark one end of each of these two lines at -3 and 1 with a small open circle to show that the points themselves are excluded.