Question 80807
Hmmm...I get something different:
Solve for x:
{{{y = x^2+2x-3}}} Set y = 0 and solve for the x's.
{{{x^2+2x-3 = 0}}} Factor the trinomial.
{{{(x+3)(x-1) = 0}}} Apply the zero products principle.
{{{x+3 = 0}}} and/or {{{x-1 = 0}}}
So...
{{{x = -3}}} and  {{{x = 1}}}

A good way to verify this is to look at the graph of the original equation:
{{{graph(300,200,-5,5,-5,5,x^2+2x-3)}}}
As you can see from the graph, the x-intercepts are:
x = -3 and x = 1