Question 536583


{{{x^2+4x+6=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+4x+6}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=4}}}, and {{{C=6}}}



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



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



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



{{{x = (-4 +- sqrt( 16-4(1)(6) ))/(2(1))}}} Square {{{4}}} to get {{{16}}}. 



{{{x = (-4 +- sqrt( 16-24 ))/(2(1))}}} Multiply {{{4(1)(6)}}} to get {{{24}}}



{{{x = (-4 +- sqrt( -8 ))/(2(1))}}} Subtract {{{24}}} from {{{16}}} to get {{{-8}}}



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



{{{x = (-4 +- 2i*sqrt(2))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



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



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



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



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



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