Question 624550


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



Notice that the quadratic {{{2m^2+2m-12}}} is in the form of {{{Am^2+Bm+C}}} where {{{A=2}}}, {{{B=2}}}, and {{{C=-12}}}



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



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



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



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



{{{m = (-2 +- sqrt( 4--96 ))/(2(2))}}} Multiply {{{4(2)(-12)}}} to get {{{-96}}}



{{{m = (-2 +- sqrt( 4+96 ))/(2(2))}}} Rewrite {{{sqrt(4--96)}}} as {{{sqrt(4+96)}}}



{{{m = (-2 +- sqrt( 100 ))/(2(2))}}} Add {{{4}}} to {{{96}}} to get {{{100}}}



{{{m = (-2 +- sqrt( 100 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{m = (-2 +- 10)/(4)}}} Take the square root of {{{100}}} to get {{{10}}}. 



{{{m = (-2 + 10)/(4)}}} or {{{m = (-2 - 10)/(4)}}} Break up the expression. 



{{{m = (8)/(4)}}} or {{{m =  (-12)/(4)}}} Combine like terms. 



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



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

  


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