Question 1076197

to find the zero {{{2x^2-x-1}}}, first set it equal to zero

{{{2x^2-x-1=0}}}.....you can use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ....in your case {{{a=2}}}, {{{b=-1}}}, and {{{c=-1}}}

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

{{{x = (1 +- sqrt( 1+8 ))/4 }}} 


{{{x = (1 +- sqrt( 9 ))/4 }}} 


{{{x = (1 +- 3)/4 }}} 

solutions:

{{{x = (1 + 3)/4 }}} =>{{{x = 4/4 }}} =>{{{x = 1}}} 

or

{{{x = (1 -3)/4 }}} =>{{{x = -2/4 }}} =>{{{x = -1/2}}}