Question 300631

{{{3n^2+48=0}}} Start with the given equation.



Notice that the quadratic {{{3n^2+48}}} is in the form of {{{An^2+Bn+C}}} where {{{A=3}}}, {{{B=0}}}, and {{{C=48}}}



Note: {{{3n^2+48=0}}} can be written as {{{3n^2+0n+48=0}}}



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



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



{{{n = (0 +- sqrt( (0)^2-4(3)(48) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=0}}}, and {{{C=48}}}



{{{n = (0 +- sqrt( 0-4(3)(48) ))/(2(3))}}} Square {{{0}}} to get {{{0}}}. 



{{{n = (0 +- sqrt( 0-576 ))/(2(3))}}} Multiply {{{4(3)(48)}}} to get {{{576}}}



{{{n = (0 +- sqrt( -576 ))/(2(3))}}} Subtract {{{576}}} from {{{0}}} to get {{{-576}}}



{{{n = (0 +- sqrt( -576 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{n = (0 +- 24*i)/(6)}}} Take the square root of {{{-576}}} to get {{{24*i}}}. 



{{{n = (0 + 24*i)/(6)}}} or {{{n = (-0 - 24*i)/(6)}}} Break up the expression. 



{{{n = (0)/(6) + (24*i)/(6)}}} or {{{n =  (-0)/(6) - (24*i)/(6)}}} Break up the fraction for each case. 



{{{n = 0+4*i}}} or {{{n =  0-4*i}}} Reduce. 



{{{n = 4*i}}} or {{{n = -4*i}}} Simplify. 



So the solutions are {{{n = 4*i}}} or {{{n = -4*i}}}