Question 352771


{{{x^2-x+11=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-1}}}, and {{{c=11}}}



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)(11) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-1}}}, and {{{c=11}}}



{{{x = (1 +- sqrt( (-1)^2-4(1)(11) ))/(2(1))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(1)(11) ))/(2(1))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-44 ))/(2(1))}}} Multiply {{{4(1)(11)}}} to get {{{44}}}



{{{x = (1 +- sqrt( -43 ))/(2(1))}}} Subtract {{{44}}} from {{{1}}} to get {{{-43}}}



{{{x = (1 +- sqrt( -43 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (1 +- i*sqrt(43))/(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(43))/(2)}}} or {{{x = (1-i*sqrt(43))/(2)}}} Break up the expression.  



So the answers are {{{x = (1+i*sqrt(43))/(2)}}} or {{{x = (1-i*sqrt(43))/(2)}}}