Question 537294


{{{m^2-m-6=0}}} Start with the given equation.



Notice that the quadratic {{{m^2-m-6}}} is in the form of {{{Am^2+Bm+C}}} where {{{A=1}}}, {{{B=-1}}}, and {{{C=-6}}}



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



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



{{{m = (-(-1) +- sqrt( (-1)^2-4(1)(-6) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-1}}}, and {{{C=-6}}}



{{{m = (1 +- sqrt( (-1)^2-4(1)(-6) ))/(2(1))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{m = (1 +- sqrt( 1-4(1)(-6) ))/(2(1))}}} Square {{{-1}}} to get {{{1}}}. 



{{{m = (1 +- sqrt( 1--24 ))/(2(1))}}} Multiply {{{4(1)(-6)}}} to get {{{-24}}}



{{{m = (1 +- sqrt( 1+24 ))/(2(1))}}} Rewrite {{{sqrt(1--24)}}} as {{{sqrt(1+24)}}}



{{{m = (1 +- sqrt( 25 ))/(2(1))}}} Add {{{1}}} to {{{24}}} to get {{{25}}}



{{{m = (1 +- sqrt( 25 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{m = (1 +- 5)/(2)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{m = (1 + 5)/(2)}}} or {{{m = (1 - 5)/(2)}}} Break up the expression. 



{{{m = (6)/(2)}}} or {{{m =  (-4)/(2)}}} Combine like terms. 



{{{m = 3}}} or {{{m = -2}}} Simplify. 



So the solutions are {{{m = 3}}} or {{{m = -2}}} 



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