Question 625687


{{{x^2-3x-1=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-3x-1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-3}}}, and {{{C=-1}}}



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



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



{{{x = (-(-3) +- sqrt( (-3)^2-4(1)(-1) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-3}}}, and {{{C=-1}}}



{{{x = (3 +- sqrt( (-3)^2-4(1)(-1) ))/(2(1))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(1)(-1) ))/(2(1))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9--4 ))/(2(1))}}} Multiply {{{4(1)(-1)}}} to get {{{-4}}}



{{{x = (3 +- sqrt( 9+4 ))/(2(1))}}} Rewrite {{{sqrt(9--4)}}} as {{{sqrt(9+4)}}}



{{{x = (3 +- sqrt( 13 ))/(2(1))}}} Add {{{9}}} to {{{4}}} to get {{{13}}}



{{{x = (3 +- sqrt( 13 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (3+sqrt(13))/(2)}}} or {{{x = (3-sqrt(13))/(2)}}} Break up the expression.  



So the solutions are {{{x = (3+sqrt(13))/(2)}}} or {{{x = (3-sqrt(13))/(2)}}} 


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