Question 416794


{{{3x^2=2x+21}}} Start with the given equation.



{{{3x^2-2x-21=0}}} Get every term to the left side.



Notice that the quadratic {{{3x^2-2x-21}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=-2}}}, and {{{C=-21}}}



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



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



{{{x = (-(-2) +- sqrt( (-2)^2-4(3)(-21) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-2}}}, and {{{C=-21}}}



{{{x = (2 +- sqrt( (-2)^2-4(3)(-21) ))/(2(3))}}} Negate {{{-2}}} to get {{{2}}}. 



{{{x = (2 +- sqrt( 4-4(3)(-21) ))/(2(3))}}} Square {{{-2}}} to get {{{4}}}. 



{{{x = (2 +- sqrt( 4--252 ))/(2(3))}}} Multiply {{{4(3)(-21)}}} to get {{{-252}}}



{{{x = (2 +- sqrt( 4+252 ))/(2(3))}}} Rewrite {{{sqrt(4--252)}}} as {{{sqrt(4+252)}}}



{{{x = (2 +- sqrt( 256 ))/(2(3))}}} Add {{{4}}} to {{{252}}} to get {{{256}}}



{{{x = (2 +- sqrt( 256 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (2 +- 16)/(6)}}} Take the square root of {{{256}}} to get {{{16}}}. 



{{{x = (2 + 16)/(6)}}} or {{{x = (2 - 16)/(6)}}} Break up the expression. 



{{{x = (18)/(6)}}} or {{{x =  (-14)/(6)}}} Combine like terms. 



{{{x = 3}}} or {{{x = -7/3}}} Simplify. 



So the solutions are {{{x = 3}}} or {{{x = -7/3}}} 

  

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