Question 318434
{{{3x^2=2(x+1)}}} Start with the given equation.



{{{3x^2=2x+2}}} Distribute.



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



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



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)(-2) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-2}}}, and {{{C=-2}}}



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



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



{{{x = (2 +- sqrt( 4--24 ))/(2(3))}}} Multiply {{{4(3)(-2)}}} to get {{{-24}}}



{{{x = (2 +- sqrt( 4+24 ))/(2(3))}}} Rewrite {{{sqrt(4--24)}}} as {{{sqrt(4+24)}}}



{{{x = (2 +- sqrt( 28 ))/(2(3))}}} Add {{{4}}} to {{{24}}} to get {{{28}}}



{{{x = (2 +- sqrt( 28 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (2 +- 2*sqrt(7))/(6)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (2+2*sqrt(7))/(6)}}} or {{{x = (2-2*sqrt(7))/(6)}}} Break up the expression.  



{{{x = (1+sqrt(7))/(3)}}} or {{{x = (1-sqrt(7))/(3)}}} Reduce.



So the solutions are {{{x = (1+sqrt(7))/(3)}}} or {{{x = (1-sqrt(7))/(3)}}}