Question 846997
This is quite the complex problem!  So, let's get started.


If we rewrite {{{x^4+x^2+1=0}}} in the form of a quadratic equation, we will be able to find our zeroes.  We can do this by rewriting as:


{{{(x^2)^2+(x)x+1=0}}}


This puts our equation in standard quadratic form of ax^2+bx+c=0.  Now, we can replace the first x inside the parenthesis with a letter and replace the second x on the outside of the parenthesis with the same letter.  Let's use u:


{{{u^2+u+1=0}}}


Now we can solve using the quadratic formula.  This will give us:


{{{u=(-1 +- sqrt( -3 ))/(2) }}}


Now, here comes the tricky part.  We have to replace our u with our original variable and power:  x^2, which will turn this into:


{{{x^2=(-1 +- sqrt( -3 ))/(2) }}}


Because we want the value of x and not x^2, we will have to take the square root of both sides.  Remember, when we take the square root of both sides of an equation, we have to add the +- sign in front.  Doing this is going to give us 4 zeros:


{{{x= +- (sqrt((-1 +- sqrt( -3 ))/(2)))}}}


Next, you will want to convert the square root of -3 into an imaginary:


{{{x= +- (sqrt((-1 +- sqrt( 3 )*i)/(2)))}}}


Finally, you will need to rationalize the denominator since the denominator of the fraction is the square root of 2.  This will give you your final answer:


{{{x= +- (sqrt((2(-1 +- sqrt( 3 )*i)))/(2))}}}


For some reason, the answer typed out the word "PLUS_MINNUS-" instead of "+-".  Just replace the "PLUS_MINNUS-" with the plus/minus symbol (+-).

NOTE:  You will notice the numerator of the fraction contains a square root inside of a square root (nested root).  You will more than likely not be required to denest this root, so it was not denested in this solution.