Question 1009928
you are correct:
 {{{highlight(x=4)}}}, {{{highlight(x=3)}}}, {{{highlight(x=-1/2)}}}, {{{highlight(x=2/3)}}} are real solutions

when you factor completely your equation, you get

{{{(x-4) (x+3) (2x+1) (3x-2) (x^2-2x+2) = 0}}}

and you have two more complex solutions when {{{(x^2-2x+2) = 0}}} and they are:

{{{x^2-2x+2 = 0}}}...use quadratic formula

{{{x = (-(-2) +- sqrt( (-2)^2-4*1*2 ))/(2*1) }}} 

{{{x = (2 +- sqrt( 4-8 ))/2 }}}

{{{x = (2 +- sqrt( -4 ))/2 }}}

{{{x = (2 +- 2i)/2 }}}

{{{x = (1 +- i) }}}

solutions:

{{{highlight(x = 1-i)}}}

{{{highlight(x = 1+i)}}}