Question 194069
{{{3n^2+6=11n}}} Start with the given equation.



{{{3n^2+6-11n=0}}} Subtract 11n from both sides.



{{{3n^2-11n+6=0}}} Rearrange the terms.



Notice we have a quadratic equation in the form of {{{an^2+bn+c}}} where {{{a=3}}}, {{{b=-11}}}, and {{{c=6}}}



Let's use the quadratic formula to solve for n



{{{n = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{n = (-(-11) +- sqrt( (-11)^2-4(3)(6) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=-11}}}, and {{{c=6}}}



{{{n = (11 +- sqrt( (-11)^2-4(3)(6) ))/(2(3))}}} Negate {{{-11}}} to get {{{11}}}. 



{{{n = (11 +- sqrt( 121-4(3)(6) ))/(2(3))}}} Square {{{-11}}} to get {{{121}}}. 



{{{n = (11 +- sqrt( 121-72 ))/(2(3))}}} Multiply {{{4(3)(6)}}} to get {{{72}}}



{{{n = (11 +- sqrt( 49 ))/(2(3))}}} Subtract {{{72}}} from {{{121}}} to get {{{49}}}



{{{n = (11 +- sqrt( 49 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{n = (11 +- 7)/(6)}}} Take the square root of {{{49}}} to get {{{7}}}. 



{{{n = (11 + 7)/(6)}}} or {{{n = (11 - 7)/(6)}}} Break up the expression. 



{{{n = (18)/(6)}}} or {{{n =  (4)/(6)}}} Combine like terms. 



{{{n = 3}}} or {{{n = 2/3}}} Simplify. 



So the answers are {{{n = 3}}} or {{{n = 2/3}}}