Question 625211
{{{ 2x^2 + 11x - 1 = 0 }}}
First add {{{ 1 }}} to both sides
{{{ 2x^2 + 11x = 1 }}}
Now divide both sides by {{{ 2 }}}
{{{ x^2 + (11/2)*x = 1/2 }}}
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In order to complete the square, take
1/2 of the coefficient of the {{{ x }}} term,
square it, then add it to both sides
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{{{ x^2 + (11/2)*x + (11/4)^2 = 1/2 + (11/4)^2 }}} 
{{{ x^2 + (11/2)*x + 121/16 = 8/16 + 121/16 }}}
{{{ ( x + 11/4 )^2 = 129/16 }}}
Take the square root of both sides
{{{ x + 11/4 = sqrt(129)/4 }}}
{{{ x = ( -11 + sqrt(129) ) / 4 }}}
and
{{{ x = ( -11 - sqrt(129) ) / 4 }}}
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check:
{{{ 2x^2 + 11x - 1 = 0 }}}
{{{ 2*( -11 - sqrt(129 ) / 4 )^2 + 11*( -11 - sqrt(129) ) / 4 ) - 1 = 0 }}}
{{{ (1/8)*( 121 + 22*sqrt(129) + 129 ) - 121/4 - (11/4)*sqrt(129) - 1 = 0 }}}
{{{ 121/8 + (22/8)*sqrt(129) + 129/8 - 242/8 - (22/8)*sqrt(129) - 8/8 = 0 }}}
{{{ 121/8 + 129/8 -242/8 - 8/8 = 0 }}}
{{{ 250/8 - 250/8 = 0 }}}
{{{ 0 = 0 }}}
OK
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