Question 245751
Just treat this quadratic as you normally would, but make sure to follow the rules of complex arithmetic.



{{{x^2+2ix-3=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+2ix-3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=2i}}}, and {{{C=-3}}}



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



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



{{{x = (-(2i) +- sqrt( (2i)^2-4(1)(-3) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=2i}}}, and {{{C=-3}}}



{{{x = (-2i +- sqrt( -4-4(1)(-3) ))/(2(1))}}} Square {{{2i}}} to get {{{(2i)^2=(2i)(2i)=4i^2=4(-1)=-4}}}



Note: Since {{{i=sqrt(-1)}}}, {{{i^2=(sqrt(-1))^2=-1}}}



{{{x = (-2i +- sqrt( -4+12 ))/(2(1))}}} Multiply -4, 1 and -3 to get {{{-4(1)(-3)=12}}}



{{{x = (-2i +- sqrt( -4+12 ))/(2)}}} Multiply 2 and 1 to get 2.



{{{x = (-2i +- sqrt( 8 ))/(2)}}} Combine like terms.



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



{{{x = -i +- sqrt( 2 )}}} Reduce.



{{{x = -i + sqrt( 2 )}}} or {{{x = -i - sqrt( 2 )}}} Break up the 'plus/minus'



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