Question 712491
{{{f(x)= 2x^2-x-1}}}

If {{{f(x)=1}}},then

{{{1= 2x^2-x-1}}}

or

{{{2x^2-x-1=1}}}

{{{2x^2-x-1-1=0}}}

{{{2x^2-x-2=0}}}...use quadratic formula to solve for {{{x}}}


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

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

 {{{x = (1 +- sqrt( 1+16 ))/4 }}}

 {{{x = (1 +- sqrt( 17 ))/4 }}}

 {{{x = (1 +- 4.12)/4 }}}

solutions:

 {{{x = (1 +4.12)/4 }}}

 {{{x = 5.12/4 }}}

 {{{x = 1.28 }}}

or

{{{x = (1 -4.12)/4 }}}

 {{{x = -3.12/4 }}}

 {{{x =-0.78 }}}




{{{drawing( 600, 600, -5, 5, -5, 5,grid(1),circle(-0.78,1,0.1),locate(-1.5,1-.2,"(-0.78,1)"),
locate(1.28,1-.2,"(1.28,1)"),circle(1.28,1,0.1),graph( 600, 600, -5, 5, -5, 5,2x^2-x-1,1))}}}