Question 568662


{{{3x^2-x-5=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2-x-5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=-1}}}, and {{{C=-5}}}



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



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



{{{x = (-(-1) +- sqrt( (-1)^2-4(3)(-5) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-1}}}, and {{{C=-5}}}



{{{x = (1 +- sqrt( (-1)^2-4(3)(-5) ))/(2(3))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(3)(-5) ))/(2(3))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1--60 ))/(2(3))}}} Multiply {{{4(3)(-5)}}} to get {{{-60}}}



{{{x = (1 +- sqrt( 1+60 ))/(2(3))}}} Rewrite {{{sqrt(1--60)}}} as {{{sqrt(1+60)}}}



{{{x = (1 +- sqrt( 61 ))/(2(3))}}} Add {{{1}}} to {{{60}}} to get {{{61}}}



{{{x = (1 +- sqrt( 61 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (1+sqrt(61))/(6)}}} or {{{x = (1-sqrt(61))/(6)}}} Break up the expression.  



So the solutions are {{{x = (1+sqrt(61))/(6)}}} or {{{x = (1-sqrt(61))/(6)}}} 


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