Question 44074
rearrang the equation so you have zero on one side of the "=" sign.
This should give {{{x^2+2x-3=0}}}
NOw use the quadratic solver {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
a = the amount of x squared's = 1
b = the amount of x's = 2
c = the number at the end = -3
put these values in the solver to get 
{{{x = (-2 +- sqrt( 2^2-4*1*(-3) ))/(2*1) }}} 
{{{x = (-2 +- sqrt( 4+12 ))/2 }}}
{{{x = (-2 +- sqrt( 16 ))/2 }}} 
{{{x = (-2 +- 4)/2 }}} 
{{{x = -1 +- 2 }}} 
Therefore x=-3 or x=1

I hope this helps

P.S. I am trying to start up my own homework help website. I would be extremely grateful if you would e-mail me some feedback on the help you recieved to adam.chapman@student.manchester.ac.uk