```Question 502785

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 = (-(-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}}}
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