```Question 262441

Notice that the quadratic {{{2x^2-6x+3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=-6}}}, and {{{C=3}}}

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

{{{x = (-(-6) +- sqrt( (-6)^2-4(2)(3) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=-6}}}, and {{{C=3}}}

{{{x = (6 +- sqrt( (-6)^2-4(2)(3) ))/(2(2))}}} Negate {{{-6}}} to get {{{6}}}.

{{{x = (6 +- sqrt( 36-4(2)(3) ))/(2(2))}}} Square {{{-6}}} to get {{{36}}}.

{{{x = (6 +- sqrt( 36-24 ))/(2(2))}}} Multiply {{{4(2)(3)}}} to get {{{24}}}

{{{x = (6 +- sqrt( 12 ))/(2(2))}}} Subtract {{{24}}} from {{{36}}} to get {{{12}}}

{{{x = (6 +- sqrt( 12 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}.

{{{x = (6 +- 2*sqrt(3))/(4)}}} 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 = (6+2*sqrt(3))/(4)}}} or {{{x = (6-2*sqrt(3))/(4)}}} Break up the expression.

{{{x = (3+sqrt(3))/(2)}}} or {{{x = (3-sqrt(3))/(2)}}} Reduce.

So the solutions are {{{x = (3+sqrt(3))/(2)}}} or {{{x = (3-sqrt(3))/(2)}}}  ```