Question 242799
{{{(3m-6)^2 + 3(3m-6)-10=0}}} Start with the given equation.



Let {{{z=3m-6}}}



{{{z^2 + 3z-10=0}}} Replace each '3m-6' with 'z'



Notice that the quadratic {{{z^2+3z-10}}} is in the form of {{{Az^2+Bz+C}}} where {{{A=1}}}, {{{B=3}}}, and {{{C=-10}}}



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



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



{{{z = (-(3) +- sqrt( (3)^2-4(1)(-10) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=3}}}, and {{{C=-10}}}



{{{z = (-3 +- sqrt( 9-4(1)(-10) ))/(2(1))}}} Square {{{3}}} to get {{{9}}}. 



{{{z = (-3 +- sqrt( 9--40 ))/(2(1))}}} Multiply {{{4(1)(-10)}}} to get {{{-40}}}



{{{z = (-3 +- sqrt( 9+40 ))/(2(1))}}} Rewrite {{{sqrt(9--40)}}} as {{{sqrt(9+40)}}}



{{{z = (-3 +- sqrt( 49 ))/(2(1))}}} Add {{{9}}} to {{{40}}} to get {{{49}}}



{{{z = (-3 +- sqrt( 49 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{z = (-3 +- 7)/(2)}}} Take the square root of {{{49}}} to get {{{7}}}. 



{{{z = (-3 + 7)/(2)}}} or {{{z = (-3 - 7)/(2)}}} Break up the expression. 



{{{z = (4)/(2)}}} or {{{z =  (-10)/(2)}}} Combine like terms. 



{{{z = 2}}} or {{{z = -5}}} Simplify. 



So the solutions in terms of z are {{{z = 2}}} or {{{z = -5}}} 

  

Since {{{z=3m-6}}}, this means that {{{3m-6=2}}} or {{{3m-6=-5}}}. Solving for 'm' gets us the final solutions



{{{m=8/3}}} or {{{m=1/3}}}