Question 502785


{{{3n^2-2n-56=0}}} Start with the given equation.



Notice that the quadratic {{{3n^2-2n-56}}} is in the form of {{{An^2+Bn+C}}} where {{{A=3}}}, {{{B=-2}}}, and {{{C=-56}}}



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



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



{{{n = (-(-2) +- sqrt( (-2)^2-4(3)(-56) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-2}}}, and {{{C=-56}}}



{{{n = (2 +- sqrt( (-2)^2-4(3)(-56) ))/(2(3))}}} Negate {{{-2}}} to get {{{2}}}. 



{{{n = (2 +- sqrt( 4-4(3)(-56) ))/(2(3))}}} Square {{{-2}}} to get {{{4}}}. 



{{{n = (2 +- sqrt( 4--672 ))/(2(3))}}} Multiply {{{4(3)(-56)}}} to get {{{-672}}}



{{{n = (2 +- sqrt( 4+672 ))/(2(3))}}} Rewrite {{{sqrt(4--672)}}} as {{{sqrt(4+672)}}}



{{{n = (2 +- sqrt( 676 ))/(2(3))}}} Add {{{4}}} to {{{672}}} to get {{{676}}}



{{{n = (2 +- sqrt( 676 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{n = (2 +- 26)/(6)}}} Take the square root of {{{676}}} to get {{{26}}}. 



{{{n = (2 + 26)/(6)}}} or {{{n = (2 - 26)/(6)}}} Break up the expression. 



{{{n = (28)/(6)}}} or {{{n =  (-24)/(6)}}} Combine like terms. 



{{{n = 14/3}}} or {{{n = -4}}} Simplify. 



So the solutions are {{{n = 14/3}}} or {{{n = -4}}}