Question 392966


{{{x^2+6x-3=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+6x-3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=6}}}, 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 = (-(6) +- sqrt( (6)^2-4(1)(-3) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=6}}}, and {{{C=-3}}}



{{{x = (-6 +- sqrt( 36-4(1)(-3) ))/(2(1))}}} Square {{{6}}} to get {{{36}}}. 



{{{x = (-6 +- sqrt( 36--12 ))/(2(1))}}} Multiply {{{4(1)(-3)}}} to get {{{-12}}}



{{{x = (-6 +- sqrt( 36+12 ))/(2(1))}}} Rewrite {{{sqrt(36--12)}}} as {{{sqrt(36+12)}}}



{{{x = (-6 +- sqrt( 48 ))/(2(1))}}} Add {{{36}}} to {{{12}}} to get {{{48}}}



{{{x = (-6 +- sqrt( 48 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-6 +- 4*sqrt(3))/(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) +- (4*sqrt(3))/(2)}}} Break up the fraction.  



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



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



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



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Jim