```Question 409257

Notice we have a quadratic equation in the form of {{{an^2+bn+c}}} where {{{a=3}}}, {{{b=-2}}}, and {{{c=5}}}

Let's use the quadratic formula to solve for n

{{{n = (-(-2) +- sqrt( (-2)^2-4(3)(5) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=-2}}}, and {{{c=5}}}

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

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

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

{{{n = (2 +- sqrt( -56 ))/(2(3))}}} Subtract {{{60}}} from {{{4}}} to get {{{-56}}}

{{{n = (2 +- sqrt( -56 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}.

{{{n = (2 +- 2i*sqrt(14))/(6)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)

{{{n = (2+2i*sqrt(14))/(6)}}} or {{{n = (2-2i*sqrt(14))/(6)}}} Break up the expression.

{{{n = (1+i*sqrt(14))/(3)}}} or {{{n = (1-i*sqrt(14))/(3)}}} Reduce

So the answers are {{{n = (1+i*sqrt(14))/(3)}}} or {{{n = (1-i*sqrt(14))/(3)}}}

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Jim```