Question 706469


clearly in this space {{{p^2+p=1}}}.... you solve for {{{p}}} 


{{{p^2+p-1=0}}}....use quadratic formula


{{{p= (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ...note that {{{a=1}}}, {{{b=1}}} and {{{c=-1}}}


{{{p= (-1 +- sqrt( 1^2-4*1*(-1) ))/(2*1) }}}


{{{p= (-1 +- sqrt( 1+4 ))/2 }}}


{{{p= (-1 +- sqrt( 5 ))/2 }}}


{{{p= (-1 +-2.24)/2 }}}

solution:


{{{p= (-1 +2.24)/2 }}}

{{{p= (1.24)/2 }}}

{{{p= 0.62 }}}