Question 540200


{{{x^2+6x+41=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+6x+41}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=6}}}, and {{{C=41}}}



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)(41) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=6}}}, and {{{C=41}}}



{{{x = (-6 +- sqrt( 36-4(1)(41) ))/(2(1))}}} Square {{{6}}} to get {{{36}}}. 



{{{x = (-6 +- sqrt( 36-164 ))/(2(1))}}} Multiply {{{4(1)(41)}}} to get {{{164}}}



{{{x = (-6 +- sqrt( -128 ))/(2(1))}}} Subtract {{{164}}} from {{{36}}} to get {{{-128}}}



{{{x = (-6 +- sqrt( -128 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-6 +- 8i*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 = (-6)/(2) +- (8i*sqrt(2))/(2)}}} Break up the fraction.  



{{{x = -3 +- 4*sqrt(2)*i}}} Reduce.  



{{{x = -3+4*sqrt(2)*i}}} or {{{x = -3-4*sqrt(2)*i}}} Break up the expression.  



So the solutions are {{{x = -3+4*sqrt(2)*i}}} or {{{x = -3-4*sqrt(2)*i}}} 



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com">jim_thompson5910@hotmail.com</a>


Also, please consider visiting my website: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a> and making a donation. Thank you


Jim