```Question 201981

{{{x^2-x-12=0}}} Get all terms to the left side.

Notice that the quadratic {{{x^2-x-12}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-1}}}, and {{{C=-12}}}

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

{{{x = (-(-1) +- sqrt( (-1)^2-4(1)(-12) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-1}}}, and {{{C=-12}}}

{{{x = (1 +- sqrt( (-1)^2-4(1)(-12) ))/(2(1))}}} Negate {{{-1}}} to get {{{1}}}.

{{{x = (1 +- sqrt( 1-4(1)(-12) ))/(2(1))}}} Square {{{-1}}} to get {{{1}}}.

{{{x = (1 +- sqrt( 1--48 ))/(2(1))}}} Multiply {{{4(1)(-12)}}} to get {{{-48}}}

{{{x = (1 +- sqrt( 1+48 ))/(2(1))}}} Rewrite {{{sqrt(1--48)}}} as {{{sqrt(1+48)}}}

{{{x = (1 +- sqrt( 49 ))/(2(1))}}} Add {{{1}}} to {{{48}}} to get {{{49}}}

{{{x = (1 +- sqrt( 49 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}.

{{{x = (1 +- 7)/(2)}}} Take the square root of {{{49}}} to get {{{7}}}.

{{{x = (1 + 7)/(2)}}} or {{{x = (1 - 7)/(2)}}} Break up the expression.

{{{x = (8)/(2)}}} or {{{x =  (-6)/(2)}}} Combine like terms.

{{{x = 4}}} or {{{x = -3}}} Simplify.

So the solutions are {{{x = 4}}} or {{{x = -3}}}

Simply add the solutions to get {{{4+(-3)=1}}}

So the sum of the solutions is {{{1}}}```