Question 1030189
<pre>
We draw the graph of {{{y=x^2-2x-3}}} by plotting points:

(-2,5), (-1,0), (0,-3), (1,-4), (2,-3), (3,0), (4,5)

{{{drawing(3200/13,400,-3,5,-5,8,

circle(-2,5,0.15),circle(-2,5,0.13),circle(-2,5,0.11),circle(-2,5,0.09),circle(-2,5,0.07),circle(-2,5,0.05),circle(-2,5,0.03),circle(-2,5,0.01),

circle(-1,0,0.15),

circle(0,-3,0.15),circle(0,-3,0.13),circle(0,-3,0.11),circle(0,-3,0.09),circle(0,-3,0.07),circle(0,-3,0.05),circle(0,-3,0.03),circle(0,-3,0.01),

circle(1,-4,0.15),circle(1,-4,0.13),circle(1,-4,0.11),circle(1,-4,0.09),circle(1,-4,0.07),circle(1,-4,0.05),circle(1,-4,0.03),circle(1,-4,0.01),

circle(2,-3,0.15),circle(2,-3,0.13),circle(2,-3,0.11),circle(2,-3,0.09),circle(2,-3,0.07),circle(2,-3,0.05),circle(2,-3,0.03),circle(2,-3,0.01),

circle(3,0,0.15),

circle(4,5,0.15),circle(4,5,0.13),circle(4,5,0.11),circle(4,5,0.09),circle(4,5,0.07),circle(4,5,0.05),circle(4,5,0.03),circle(4,5,0.01),




graph(3200/13,400,-3,5,-5,8,(x^2-2x-3)*sqrt(-1-x)/sqrt(-1-x)),

graph(3200/13,400,-3,5,-5,8,(x^2-2x-3)*sqrt(x-3)/sqrt(x-3)),

graph(3200/13,400,-3,5,-5,8,26,(x^2-2x-3)*(sqrt(x+1)/sqrt(1+x))*(sqrt(3-x)/sqrt(3-x)))



 )}}} 

Since we want to solve {{{x^2-2x-3>0}}}, and we have the 
graph of {{{y=x^2-2x-3}}}, so we want to know where y is
greater than 0.  So the solution will be the values of x
fro which the graph is ABOVE the x-axis.  Not including
the intercepts, the red part of the graph is ABOVE the 
x-axis and the green part is below the x-axis. So the part
of the x-axis that the red part of the graph is above,
is given interval notation as {{{matrix(1,3,

(matrix(1,3,-infinity,",",-1)),
U,
(matrix(1,3,3,",",infinity))  

)}}}

Edwin</pre>