Question 200553

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



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



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)(2) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=1}}}, and {{{C=2}}}



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



{{{x = (-1 +- sqrt( 1-8 ))/(2(1))}}} Multiply {{{4(1)(2)}}} to get {{{8}}}



{{{x = (-1 +- sqrt( -7 ))/(2(1))}}} Subtract {{{8}}} from {{{1}}} to get {{{-7}}}



{{{x = (-1 +- sqrt( -7 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-1 +- i*sqrt(7))/(2)}}} Simplify the square root  



{{{x = (-1+i*sqrt(7))/(2)}}} or {{{x = (-1-i*sqrt(7))/(2)}}} Break up the expression.  



So the solutions are {{{x = (-1+i*sqrt(7))/(2)}}} or {{{x = (-1-i*sqrt(7))/(2)}}}