Question 473450
{{{n(n+2)=143}}} Start with the given equation



{{{n^2+2n=143}}} Distribute



{{{n^2+2n-143=0}}} Subtract 143 from both sides.



Notice that the quadratic {{{n^2+2n-143}}} is in the form of {{{An^2+Bn+C}}} where {{{A=1}}}, {{{B=2}}}, and {{{C=-143}}}



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(1)(-143) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=2}}}, and {{{C=-143}}}



{{{n = (-2 +- sqrt( 4-4(1)(-143) ))/(2(1))}}} Square {{{2}}} to get {{{4}}}. 



{{{n = (-2 +- sqrt( 4--572 ))/(2(1))}}} Multiply {{{4(1)(-143)}}} to get {{{-572}}}



{{{n = (-2 +- sqrt( 4+572 ))/(2(1))}}} Rewrite {{{sqrt(4--572)}}} as {{{sqrt(4+572)}}}



{{{n = (-2 +- sqrt( 576 ))/(2(1))}}} Add {{{4}}} to {{{572}}} to get {{{576}}}



{{{n = (-2 +- sqrt( 576 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{n = (-2 +- 24)/(2)}}} Take the square root of {{{576}}} to get {{{24}}}. 



{{{n = (-2 + 24)/(2)}}} or {{{n = (-2 - 24)/(2)}}} Break up the expression. 



{{{n = (22)/(2)}}} or {{{n =  (-26)/(2)}}} Combine like terms. 



{{{n = 11}}} or {{{n = -13}}} Simplify. 



So the solutions are {{{n = 11}}} or {{{n = -13}}}