Question 591815


{{{x^2+6x+34=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+6x+34}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=6}}}, and {{{C=34}}}



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



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



{{{x = (-(6) +- sqrt( (6)^2-4(1)(34) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=6}}}, and {{{C=34}}}



{{{x = (-6 +- sqrt( 36-4(1)(34) ))/(2(1))}}} Square {{{6}}} to get {{{36}}}. 



{{{x = (-6 +- sqrt( 36-136 ))/(2(1))}}} Multiply {{{4(1)(34)}}} to get {{{136}}}



{{{x = (-6 +- sqrt( -100 ))/(2(1))}}} Subtract {{{136}}} from {{{36}}} to get {{{-100}}}



{{{x = (-6 +- sqrt( -100 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-6 +- 10*i)/(2)}}} Take the square root of {{{-100}}} to get {{{10*i}}}. 



{{{x = (-6 + 10*i)/(2)}}} or {{{x = (-6 - 10*i)/(2)}}} Break up the expression. 



{{{x = (-6)/(2) + (10*i)/(2)}}} or {{{x =  (-6)/(2) - (10*i)/(2)}}} Break up the fraction for each case. 



{{{x = -3+5*i}}} or {{{x =  -3-5*i}}} Reduce. 



So the solutions are {{{x = -3+5*i}}} or {{{x = -3-5*i}}}