```Question 568741

{{{x^2+3x+8-5=0}}} Subtract 5 from both sides.

{{{x^2+3x+3=0}}} Combine like terms.

Notice that the quadratic {{{x^2+3x+3}}} is 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( 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 solutions are {{{x = (-3+i*sqrt(3))/(2)}}} or {{{x = (-3-i*sqrt(3))/(2)}}}

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