Question 354990
{{{x^2+x+1=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+x+1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=1}}}, and {{{C=1}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(1) +- sqrt( (1)^2-4(1)(1) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=1}}}, and {{{C=1}}}



{{{x = (-1 +- sqrt( 1-4(1)(1) ))/(2(1))}}} Square {{{1}}} to get {{{1}}}. 



{{{x = (-1 +- sqrt( 1-4 ))/(2(1))}}} Multiply {{{4(1)(1)}}} to get {{{4}}}



{{{x = (-1 +- sqrt( -3 ))/(2(1))}}} Subtract {{{4}}} from {{{1}}} to get {{{-3}}}



{{{x = (-1 +- sqrt( -3 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-1 +- i*sqrt(3))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (-1+i*sqrt(3))/(2)}}} or {{{x = (-1-i*sqrt(3))/(2)}}} Break up the expression.  



So the solutions are {{{x = (-1+i*sqrt(3))/(2)}}} or {{{x = (-1-i*sqrt(3))/(2)}}}