Question 309975


{{{-x^2+x+12=0}}} Start with the given equation.



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



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



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



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



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