```Question 179592

{{{x^2-3x+3=0}}} Get all terms to the left side.

Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-3}}}, and {{{c=3}}}

Let's use the quadratic formula to solve for x

{{{x = (-(-3) +- sqrt( (-3)^2-4(1)(3) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-3}}}, and {{{c=3}}}

{{{x = (3 +- sqrt( (-3)^2-4(1)(3) ))/(2(1))}}} Negate {{{-3}}} to get {{{3}}}.

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

{{{x = (3 +- sqrt( 9-12 ))/(2(1))}}} Multiply {{{4(1)(3)}}} to get {{{12}}}

{{{x = (3 +- sqrt( -3 ))/(2(1))}}} Subtract {{{12}}} from {{{9}}} to get {{{-3}}}

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

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

So the answers are {{{x = (3+i*sqrt(3))/(2)}}} or {{{x = (3-i*sqrt(3))/(2)}}}

```