Question 255371
Since "The square of a number plus the number is 132", we know that {{{x^2+x=132}}}





{{{x^2+x=132}}} Start with the given equation.



{{{x^2+x-132=0}}} Subtract 132 from both sides.



Notice that the quadratic {{{x^2+x-132}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=1}}}, and {{{C=-132}}}



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)(-132) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=1}}}, and {{{C=-132}}}



{{{x = (-1 +- sqrt( 1-4(1)(-132) ))/(2(1))}}} Square {{{1}}} to get {{{1}}}. 



{{{x = (-1 +- sqrt( 1--528 ))/(2(1))}}} Multiply {{{4(1)(-132)}}} to get {{{-528}}}



{{{x = (-1 +- sqrt( 1+528 ))/(2(1))}}} Rewrite {{{sqrt(1--528)}}} as {{{sqrt(1+528)}}}



{{{x = (-1 +- sqrt( 529 ))/(2(1))}}} Add {{{1}}} to {{{528}}} to get {{{529}}}



{{{x = (-1 +- sqrt( 529 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-1 +- 23)/(2)}}} Take the square root of {{{529}}} to get {{{23}}}. 



{{{x = (-1 + 23)/(2)}}} or {{{x = (-1 - 23)/(2)}}} Break up the expression. 



{{{x = (22)/(2)}}} or {{{x =  (-24)/(2)}}} Combine like terms. 



{{{x = 11}}} or {{{x = -12}}} Simplify. 



So the solutions are {{{x = 11}}} or {{{x = -12}}} which means that the number is either 11 or -12