Question 498757


{{{x^2+2x+5=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+2x+5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=2}}}, and {{{C=5}}}



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



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



{{{x = (-(2) +- sqrt( (2)^2-4(1)(5) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=2}}}, and {{{C=5}}}



{{{x = (-2 +- sqrt( 4-4(1)(5) ))/(2(1))}}} Square {{{2}}} to get {{{4}}}. 



{{{x = (-2 +- sqrt( 4-20 ))/(2(1))}}} Multiply {{{4(1)(5)}}} to get {{{20}}}



{{{x = (-2 +- sqrt( -16 ))/(2(1))}}} Subtract {{{20}}} from {{{4}}} to get {{{-16}}}



{{{x = (-2 +- sqrt( -16 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-2 +- 4*i)/(2)}}} Take the square root of {{{-16}}} to get {{{4*i}}}. 



{{{x = (-2 + 4*i)/(2)}}} or {{{x = (-2 - 4*i)/(2)}}} Break up the expression. 



{{{x = (-2)/(2) + (4*i)/(2)}}} or {{{x =  (-2)/(2) - (4*i)/(2)}}} Break up the fraction for each case. 



{{{x = -1+2*i}}} or {{{x =  -1-2*i}}} Reduce. 



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