Question 990252
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Your equation is


{{{(x^2-x)^2+(x^2-x)-2}}} = {{{0}}}.


The standard and universal method for solving such equations is introducing new variable.


In our case let  y = {{{x^2-x}}}.


Then the original equation takes the form


{{{y^2 + y - 2}}} = {{{0}}}.


It is a quadratic equation.  You can easily solve it by applying the quadratic formula or the Viete's theorem.  Its roots are 


{{{y[1]}}} = -2,   {{{y[2]}}} = 1.


Now we should solve two quadratic equations to find &nbsp;<B>x</B>. &nbsp;They are


{{{x^2 - x}}} = {{{-2}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1) 


and


{{{x^2 -x}}} = {{{1}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)


The first of these two equations is 


{{{x^2 - x + 2}}} = {{{0}}}. 


It has no real solutions &nbsp;(the discriminant &nbsp;d = {{{b^2-4ac}}} = 1 - 4*2 = -7 &nbsp;is negative).


The second of these two equations is


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


It has two roots &nbsp;&nbsp;{{{x[1,2]}}} = {{{(1 +- sqrt(5))/2}}}.


<B>Answer</B>. &nbsp;The given equation has two roots: &nbsp;{{{(1+sqrt(5))/2}}} &nbsp;and &nbsp;{{{(1-sqrt(5))/2}}}.