Question 420394

{{{y^2+6y-1=0}}}....{{{a=1}}}, {{{b=6}}} and {{{c=-1}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


{{{x = (-6 +- sqrt( 6^2-4*1*(-1) ))/(2*1) }}}


{{{x = (-6 +- sqrt( 36+4 ))/2 }}}


{{{x = (-6 +- sqrt( 40 ))/2 }}}


{{{x = (-6 +- sqrt( 4*10 ))/2 }}}


{{{x = (-6 +- 2*sqrt( 10 ))/2 }}}


so, you have:


{{{x = (-6 + 2*sqrt( 10 ))/2 }}}


{{{x = -3 + sqrt( 10 ) }}}


or


{{{x = -3 - sqrt( 10 ) }}}