Question 539374


{{{x^2+6x=-10}}} Start with the given equation.



{{{x^2+6x+10=0}}} Get every term to the left side.



Notice that the quadratic {{{x^2+6x+10}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=6}}}, and {{{C=10}}}



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)(10) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=6}}}, and {{{C=10}}}



{{{x = (-6 +- sqrt( 36-4(1)(10) ))/(2(1))}}} Square {{{6}}} to get {{{36}}}. 



{{{x = (-6 +- sqrt( 36-40 ))/(2(1))}}} Multiply {{{4(1)(10)}}} to get {{{40}}}



{{{x = (-6 +- sqrt( -4 ))/(2(1))}}} Subtract {{{40}}} from {{{36}}} to get {{{-4}}}



{{{x = (-6 +- sqrt( -4 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-6 +- 2*i)/(2)}}} Take the square root of {{{-4}}} to get {{{2*i}}}. 



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



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



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



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



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