Question 264149


{{{3x^2-12x+11=0}}} Start with the given equation.



Notice that the quadratic {{{3x^2-12x+11}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=-12}}}, and {{{C=11}}}



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



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



{{{x = (-(-12) +- sqrt( (-12)^2-4(3)(11) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-12}}}, and {{{C=11}}}



{{{x = (12 +- sqrt( (-12)^2-4(3)(11) ))/(2(3))}}} Negate {{{-12}}} to get {{{12}}}. 



{{{x = (12 +- sqrt( 144-4(3)(11) ))/(2(3))}}} Square {{{-12}}} to get {{{144}}}. 



{{{x = (12 +- sqrt( 144-132 ))/(2(3))}}} Multiply {{{4(3)(11)}}} to get {{{132}}}



{{{x = (12 +- sqrt( 12 ))/(2(3))}}} Subtract {{{132}}} from {{{144}}} to get {{{12}}}



{{{x = (12 +- sqrt( 12 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (12 +- 2*sqrt(3))/(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 = (12+2*sqrt(3))/(6)}}} or {{{x = (12-2*sqrt(3))/(6)}}} Break up the expression.  



{{{x = (6+sqrt(3))/(3)}}} or {{{x = (6-sqrt(3))/(3)}}} Reduce.



So the solutions are {{{x = (6+sqrt(3))/(3)}}} or {{{x = (6-sqrt(3))/(3)}}}