Question 253100
Since "The square of a negative integer minus itself is 110", this means that {{{x^2-x=110}}}



{{{x^2-x=110}}} Start with the given equation.



{{{x^2-x-110=0}}} Subtract 110 from both sides.



Notice that the quadratic {{{x^2-x-110}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-1}}}, and {{{C=-110}}}



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



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



{{{x = (-(-1) +- sqrt( (-1)^2-4(1)(-110) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-1}}}, and {{{C=-110}}}



{{{x = (1 +- sqrt( (-1)^2-4(1)(-110) ))/(2(1))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(1)(-110) ))/(2(1))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1--440 ))/(2(1))}}} Multiply {{{4(1)(-110)}}} to get {{{-440}}}



{{{x = (1 +- sqrt( 1+440 ))/(2(1))}}} Rewrite {{{sqrt(1--440)}}} as {{{sqrt(1+440)}}}



{{{x = (1 +- sqrt( 441 ))/(2(1))}}} Add {{{1}}} to {{{440}}} to get {{{441}}}



{{{x = (1 +- sqrt( 441 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (1 +- 21)/(2)}}} Take the square root of {{{441}}} to get {{{21}}}. 



{{{x = (1 + 21)/(2)}}} or {{{x = (1 - 21)/(2)}}} Break up the expression. 



{{{x = (22)/(2)}}} or {{{x =  (-20)/(2)}}} Combine like terms. 



{{{x = 11}}} or {{{x = -10}}} Simplify. 



So the possible solutions are {{{x = 11}}} or {{{x = -10}}} 

  
  
But since we're told that the integer is negative, this means that the only solution is {{{x=-10}}}